STATICS AND
MECHANICS OF
MATERIALS
This page intentionally left blank
STATICS AND
MECHANICS OF
MATERIALS
FIFTH EDITION
R. C. HIBBELER
PEARSON
Hoboken
Boston
Columbu s
San Francisco
New York
Ind ianapol is London Toronto Sydney Singapore Tokyo Montreal Dubai
Mad rid Hong Kong Mexico City Munich Pa ris Amsterdam Cape Town
Vice President and Editorial Director, ECS: Marcia 1. Horron
Editor in Chief: 111/ian Parrridge
Senior Editor: Norrin Dias
Editorial Assistant: Michelle Bayman
Program/Project Management Team Lead: Scott Disanno
Program Manager: Sandra L. Rodriguez
Project Manager: Rose Kernan
Director of Operations: Nick Sk/irsis
Operations Specialist Ma11ra Zaldivar-Garcia
Field Marketing Manager: Deme1ri11s Hall
Marketing Assistant: Jon Bryanl
Cover Designer: Black Horse Designs
Cover Image: Adam Lister/Geny Images
Credits and acknowledgments borrowed from other sources and reproduced, with permission, in this textbook
appear on appropriate page within text. Unless otherwise specified, all photos provided by R.C. Hibbeler.
Copyright © 2017, 2014, 2011 by Pearson Education, Inc. or its affiliates. All Rights Reserved. Printed in the
United States of America. This publication is protected by copyright, and permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form
or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding
permissions, request forms and the appropriate contacts within the Pearson Education Global Rights &
Permissions Department, please visit www.pearsoned.com/permissions/.
Many of the designations by manufacturers and seller to distinguish their products are claimed as trademarks.
Where those designations appear in this book, and the publisher was aware of a trademark claim, the
designations have been printed in initial caps or all caps.
The author and publisher of this book have used their best efforts in preparing this book. These efforts include
the development, research, and testing of theories and programs to determine their effectiveness. The author
and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the
documentation contained in this book. The author and publisher shall not be liable in any event for incidental
or consequential damages with, or arising out of, the furnishing, performance, or use of these programs.
Pearson Education Ltd., London
Pearson Education Singapore, Pte. Ltd
Pearson Education Canada, Inc.
Pearson Education-Japan
Pearson Education Australia PTY, Limited
Pearson Education North Asia, Ltd., Hong Kong
Pearson Educaci6n de Mexico, S.A. de C. V.
Pearson Education Malaysia, Pte. Ltd.
Pearson Education, Inc., Hoboken, New Jersey
Library of Congress Cataloging-in-Publication Data
Hibbeler, R. C.
Statics and mechanics of materials I R.C. Hibbeler.
Fifth edition. I Hoboken : Pearson, 2016. I Includes index.
LCCN 2016013512 I ISBN 9780134382593
LCSH: Strength of materials. I Statics. I Structural analysis (Engineering)
LCC TA405 .H48 2016 I DOC 620.1112 - dc23
LC record available at http:l/lccn.loc.gov/2016013512
10987654321
PEARSON
ISBN 10: 0-13-438259-5
ISBN 13: 978-0-13-438259-3
To the Student
With the hope that this work will stimulate
an interest in Engineering Mechanics and
Mechanics of Materials and provide
an acceptable guide to its understanding.
This page intentionally left blank
PREFACE
This book represents a combined abridged version of two of the author's
books, namely Engineering Mechanics: Statics, Fourteenth Edition and
Mechanics of Materials, Tenth Edition. It provides a clear and thorough
presentation of both the theory and application of the important
fundamental topics of these subjects, that are often used in many
engineering disciplines. The development emphasizes the importance of
satisfying equilibrium, compatibility of deformation, and material
behavior requirements. The hallmark of the book, however, remains the
same as the author's unabridged versions, and that is, strong emphasis is
placed on drawing a free-body diagram, and the importance of selecting
an appropriate coordinate system and an associated sign convention
whenever the equations of mechanics are applied. Throughout the book,
many analysis and design applications are presented, which involve
mechanical elements and structural members often encountered in
engineering practice.
NEW TO THIS EDITION
• Preliminary Problems. This new feature can be found throughout the
text, and is given just before the Fundamental Problems. The intent here
is to test the student's conceptual understanding of the theory. Normally
the solutions require little or no calculation, and as such, these problems
provide a basic understanding of the concepts before they are applied
numerically. All the solutions are given in the back of the text.
• Improved Fundamental Problems. These problem sets are located just
after the Preliminary Problems. They offer students basic applications of
the concepts covered in each section, and they help provide the chance to
develop their problem-solving skills before attempting to solve any of the
standard problems that follow.
• New Problems. There are approximately 80% new problems that have
been added to this edition, which involve applications to many different
fields of engineering.
• Updated Material. Many topics in the book have been re-written in
order to further enhance clarity and to be more succinct. Also, some of
the artwork has been enlarged and improved throughout the book to
support these changes.
VI 11
PREFAC E
• New Layout Design. Additional design features have been added to
this edition to provide a better display of the material. Almost all the
topics are presented on a one or two page spread so that page turning is
minimized.
• New Photos. The relevance of knowing the subject matter is reflected by
the real-world application of new or updated photos placed throughout the
book. These photos generally are used to explain how the principles apply
to real-world situations and how materials behave under load.
HALLMARK FEATURES
Besides the new features just mentioned, other outstanding features that
define the contents of the text include the following.
Organization and Approach. Each chapter is organized into welldefined sections that contain an explanation of specific topics, illustrative
example problems, and a set of homework problems. The topics within
each section are placed into subgroups defined by boldface titles. The
purpose of this is to present a structured method for introducing each new
definition or concept and to make the book convenient for later reference
and review.
Chapter Contents. Each chapter begins with a photo demonstrating
a broad-range application of the material within the chapter. A bulleted
list of the chapter contents is provided to give a general overview of the
material that will be covered.
Emphasis on Free-Body Diagrams. Drawing a free-body
diagram is particularly important when solving problems, and for this
reason this step is strongly emphasized throughout the book. In particular,
within the statics coverage some sections are devoted to show how to
draw free-body diagrams. Specific homework problems have also been
added to develop this practice.
Procedures for Analysis. A general procedure for analyzing any
mechanics problem is presented at the end of the first chapter. Then this
procedure is customized to relate to specific types of problems that are
covered throughout the book . This unique feature provides the student
with a logical and orderly method to follow when applying the theory.The
example problems are solved using this outlined method in order to
clarify its numerical application. Realize, however, that once the relevant
principles have been mastered and enough confidence and judgment have
been obtained, the student can then develop his or her own procedures
for solving problems.
PREFACE
Important Points. This feature provides a review or summary of the
most important concepts in a section and highlights the most significant
points that should be realized when applying the theory to solve problems.
Conceptual Understanding. Through the use of photographs
placed throughout the book, the theory is applied in a simplified way in
order to illustrate some of its more important conceptual features and
instill the physical meaning of many of the terms used in the equations.
These simplified applications increase interest in the subject matter and
better prepare the student to understand the examples and solve problems.
Preliminary and Fundamental Problems. These problems may
be considered as extended examples, since the key equations and answers
are all listed in the back of the book. Additionally, when assigned, these
problems offer students an excellent means of preparing for exams, and
they can be used at a later time as a review when studying for the
Fundamentals of Engineering Exam.
Conceptual Problems. Throughout the text, usually at the end of
each chapter, there is a set of problems that involve conceptual situations
related to the application of the principles contained in the chapter. These
analysis and design problems are intended to engage students in thinking
through a real-life situation as depicted in a photo. They can be assigned
after the students have developed some expertise in the subject matter
and they work well either for individual or team projects.
Homework Problems. Apart from the Preliminary, Fundamental,
and Conceptual type problems mentioned previously, other types of
problems contained in the book include the following:
• General Analysis and Design Problems. The majority of problems
in the book depict realistic situations encountered in engineering
practice. Some of these problems come from actual products used in
industry. It is hoped that this realism will both stimulate the student's
interest in engineering mechanics and provide a means for developing
the skill to reduce any such problem from its physical description to a
model or symbolic representation to which the principles of mechanics
may be applied.
Throughout the book, there is an approximate balance of problems
using either SI of FPS units. Furthermore, in any set, an attempt has
been made to arrange the problems in order of increasing difficulty,
except for the end of chapter review problems, which are presented in
random order. Problems that are simply indicated by a problem
number have an answer given in the back of the book. However, an
asterisk (*) before every fourth problem number indicates a problem
without an answer.
IX
x
PREFACE
Accuracy. In addition to the author, the text and problem solutions
have been thoroughly checked for accuracy by four other parties: Scott
Hendricks, Virginia Polytechnic Institute and State University; Karim
Nohra, University of South Florida; Kurt Norlin, Bittner Development
Group; and finally Kai Beng Yap, a practicing engineer.
CONTENTS
The book is divided into two parts, and the material is covered in the
traditional manner.
Statics. The subject of statics is presented in 6 chapters. The text begins
in Chapter 1 with an introduction to mechanics and a discussion of units.
The notion of a vector and the properties of a concurrent force system are
introduced in Chapter 2. Chapter 3 contains a general discussion of
concentrated force systems and the methods used to simplify them. The
principles of rigid-body equililbrium are developed in Chapter 4 and then
applied to specific problems involving the equilibrium of trusses, frames,
and machines in Chapter 5. Finally, topics related to the center of gravity,
centroid, and moment of inertia are treated in Chapter 6.
Mechanics of Materials. This portion of the text is covered in
10 chapters. Chapter 7 begins with a formal definition of both normal and
shear stress, and a discussion of normal stress in axially loaded members
and average shear stress caused by direct shear; finally, normal and shear
strain are defined. In Chapter 8 a discussion of some of the important
mechanical properties of materials is given. Separate treatments of axial
load, torsion, bending, and transverse shear are presented in Chapters 9, 10,
11, and 12, respectively. Chapter 13 provides a partial review of the material
covered in the previous chapters, in which the state of stress resulting from
combined loadings is discussed. In Chapter 14 the concepts for transforming
stress and strain are presented. Chapter 15 provides a means for a further
summary and review of previous material by covering design of beams
based on allowable stress. In Chapter 16 various methods for computing
deflections of beams are presented, including the method for finding the
reactions on these members if they are statically indeterminate. Lastly,
Chapter 17 provides a discussion of column buckling.
Sections of the book that contain more advanced material are indicated
by a star (*).Time permitting, some of these topics may be included in
the course. Furthermore, this material provides a suitable reference for
basic principles when it is covered in other courses, and it can be used as
a basis for assigning special projects.
Alternative Method for Coverage of Mechanics of
Materials. It is possible to cover many of the topics in the text in several
different sequences. For example, some instructors prefer to cover stress
and strain transformations first, before discussing specific applications of
PREFACE
axial load, torsion, bending, and shear. One possible method for doing this
would be to first cover stress and strain and its transformations, Chapter 7
and Chapter 14, then Chapters 8 through 13 can be covered with no Joss in
continuity.
ACKNOWLEDGMENTS
Over the years, this text has been shaped by the suggestions and
comments of many of my colleagues in the teaching profession. Their
encouragement and willingness to provide constructive criticism are very
much appreciated and it is hoped that they will accept this anonymous
recognition. A note of thanks is also given to the reviewers of both my
Engineering Mechanics: Statics and Mechanics of Materials texts. Their
comments have guided the improvement of this book as well.
In particular, I would like to thank:
•
•
•
•
•
D. Kingsbury- Arizona State University
S. Redkar- Arizona State University
P. Mokashi- Ohio State University
S. Seetharaman- Ohio State University
H. Salim- University of Missouri- Columbia
During the production process I am thankful for the assistance of Rose
Kernan, my production editor for many years, and to my wife, Conny, for
her help in proofreading and typing, that was needed to prepare the
manuscript for publication.
I would also like to thank all my students who have used the previous
edition and have made comments to improve its contents; including
those in the teaching profession who have taken the time to e-mail me
their comments.
I would greatly appreciate hearing from you if at any time you have
any comments or suggestions regarding the contents of this edition.
Russell Charles Hibbeler
hibbeler@bellsouth.net
XI
your work...
i1Q
r
=.
-~
(l
I""" • ...,_
--
(,. ':n,
' \.
-I -
'
~ 3.~7:'.l ..... 1Y\..
,:L_
A = .s-1:0.t" x s: = C- ni" xo..l ';(i..~
I
A\
U(
,-
I
I
,,.-•
= ~A =-3 -3 ~
11'\.
')'1
11 '
I" •
'2-
)( o(_o( • (l_ t11_
1.,...
= -=tr,. '1
0- '- 'Z.,.
= :;~. c. 1n
'
in.
~
your answer specific feedback
Express your answer to three significant figures and include appropriate units.
Q=
176.9
Submit
3
__
ll~
in
Hints Mv Answers Give Up Review Part
Incorrect; Try Again; 5 attempts remaining
The distance between the horizontal centroidal axis of area A' and the neutral axis of the
beam's cross section is half the distance between the top of the shaft and the neutral axis.
www. MasteringEngineering:com
XIV
PREFACE
RESOURCES FOR INSTRUCTORS
• MasteringEngineering. This online Tutorial Homework program allows
you to integrate dynamic homework with automatic grading and adaptive
tutoring. MasteringEngineering allows you to easily track the performance
of your entire class on an assignment-by-assignment basis, or the detailed
work of an individual student.
• Instructor's Solutions Manual. An instructor's solutions manual was
prepared by the author. The manual was also checked as part of the
accuracy checking program. The Instructor Solutions Manual is available
at www.pearsonhighered.com.
• Presentation Resources. All art from the text is available in Power Point
slide and JPEG format. These files are available for download from the
Instructor Resource Center at www.pearsonhighered.com. If you are in
need of a login and password for this site, please contact your local
Pearson representative.
• Video Solutions. Developed by Professor Edward Berger, Purdue
University, video solutions located on the Pearson Engineering Portal
offer step-by-step solution walkthroughs of representative homework
problems from each section of the text. Make efficient use of class time
and office hours by showing students the complete and concise problem
solving approaches that they can access anytime and view at their own
pace. The videos are designed to be a flexible resource to be used however
each instructor and student prefers. A valuable tutorial resource, the
videos are also helpful for student self-evaluation as students can pause
the videos to check their understanding and work alongside the video.
Access the videos at pearsonhighered.com/engineering-resources/ and
follow the links for the Statics and Mechanics of Materials text.
RESOURCES FOR STUDENTS
• Mastering Engineering. Tutorial homework problems emulate the
instrutor's office-hour environment.
• Engineering Portal - The Pearson Engineering Portal, located at
pearsonhighered.com/engineering-resources/ includes opportunities for
practice and review including:
• Video Solutions- Complete, step-by-step solution walkthroughs of
representative homework problems from each section of the text. Videos
offer fully worked solutions that show every step of the representative
homework problems- this helps students make vital connections between
concepts.
CONTENTS
1
1.1
1.2
1.3
1.4
1.5
2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
General Princip les
3
Chapter Objectives 3
Mechanics 3
Fundamental Concepts 4
The International System of Units 8
Numerical Calculations 10
General Procedure for Ana lysis 11
Force Vectors
17
Chapter Objectives 17
Scalars and Vectors 17
Vector Operations 18
Vector Addition of Forces 20
Addition of a System of Coplanar Forces 31
Cartesian Vectors 40
Addition of Cartesian Vectors 43
Position Vectors 52
Force Vector Directed Along a Line 55
Dot Product 63
Force System
3
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
Resu ltants
79
Chapter Objectives 79
Moment of a Force-Scalar Formulation 79
Cross Product 83
Moment of a Force-Vector Formulation 86
Principle of Moments 90
Moment of a Force about a
Specified Axis 101
Moment of a Couple 110
Simplification of a Force and Couple
System 120
Further Simplification of a Force and
Couple System 131
Reduction of a Simple Distributed
Loading 143
4
4 .1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
Eq ui librium of a
R igid Body
157
Chapter Objectives 157
Conditions for Rigid-Body Equilibrium 157
Free-Body Diagrams 159
Equations of Equilibrium 169
Two- and Three-Force Members 175
Free-Body Diagrams 185
Equations of Equilibrium 190
Characteristics of Dry Friction 200
Problems Involving Dry Friction 204
5
Stru ctural Ana lys is
5.1
5.2
5 .3
5 .4
5 .5
Chapter Objectives 223
Simple Trusses 223
The Method of Joints 226
Zero-Force Members 232
The Method of Sections 239
Frames and Machines 248
223
Center o f G ravity,
6
6.1
6.2
6.3
6.4
6 .5
C entroid, and Moment
of Inertia
269
Chapter Objectives 269
Center of Gravity and the Centroid
of a Body 269
Composite Bodies 283
Moments of Inertia for Areas 292
Parallel-Axis Theorem for an Area 293
Moments of Inertia for Composite Areas 301
XVI
7
7 .1
7 .2
7 .3
7.4
7 .5
7.6
7.7
7 .8
CONTENTS
Stress and Strain
311
Chapter Objectives 311
Introduction 311
Internal Resultant Loadings 312
Stress 326
Average Norma l Stress in an
Axia lly Loaded Bar 328
Average Shear Stress 335
Allowable Stress Design 346
Deformation 361
Strain 362
10
10.1
10.2
10.3
10.4
10.5
11
8
8.1
8.2
8.3
8.4
8.5
8.6
Mechanical Properties
of Materials
379
Chapter Objectives 379
The Tension and Compression Test 379
The Stress- Strain Diagram 381
Stress- Strain Behavior of Ductile and
Brittle Materials 385
Strain Energy 389
Poisson's Ratio 398
The Shear Stress-Strain Diagram 400
11 .1
11 .2
11 .3
11 .4
11 .5
12
9
9.1
9.2
9.3
9.4
9.5
9.6
Axial Load
411
Chapter Objectives 411
Saint-Venant's Principle 411
Elastic Deformation of an Axia lly
Loaded Member 413
Principle of Superposition 428
Statically Indeterminate Axially Loaded
Members 428
The Force Method of Analysis for Axia lly
Loaded Members 435
Thermal Stress 441
Torsion
453
Chapter Objectives 453
Torsional Deformation of a
Circular Shaft 453
The Torsion Formula 456
Power Transmission 464
Angle of Twist 474
Statically Indeterminate Torque-Loaded
Members 488
Bending
499
Chapter Objectives 499
Shear and Moment Diagrams 499
Graphical Method for Constructing
Shear and Moment Diagrams 506
Bending Deformation of a
Straight Member 525
The Flexure Formula 529
Unsymmetric Bending 544
Transverse Shear
559
Chapter Objectives 559
12.1 Shear in Straight Members 559
12.2 The Shear Formula 560
12.3 Shear Flow in Built-Up Members 578
13
Comb ined Loadings
Chapter Objectives 591
13.1 Thin-Walled Pressure Vessels 591
13.2 State of Stress Caused by Combined
Loadings 598
591
CONTENTS
14
Stress and Strain
Transformation
619
Chapter Objectives 619
14.1
Plane-Stress Transformation 619
14.2
General Equations of Plane-Stress
Transformation 624
14.3
Principal Stresses and Maximum
In-Plane Shear Stress 627
14.4
Mohr's Circle-Plane Stress 643
14.5
Absolute Maximum Shear Stress 655
14.6
Plane Strain 661
14.7
General Equations of Plane-Strain
Transformation 662
"14.8
Mohr's Circle-Plane Strain 670
"14.9
Absolute Maximum Shear Strain 678
14.10 Strain Rosettes 680
14.11 Material Property Relationships 682
17
17.1
17.2
17.3
"17.4
Buckling of Columns
A
B
D
Chapter Objectives 777
Critical Load 777
Ideal Column with Pin Supports 780
Columns Having Various Types of
Supports 786
The Secant Formula 798
Mathematical Review and Expressions 810
Geometric Properties of An Area and
Volume 814
Geometric Properties of Wide-Flange
Sections 816
Slopes and Deflections of Beams 820
Preliminary Problems Solutions 822
Design of Beams and
Shafts
699
Chapter Objectives 699
15.1 Basis for Beam Design 699
15.2 Prismatic Beam Design 702
16
16.1
16.2
"16.3
16.4
16.5
Deflection of Beams
and Sha~s
Fundamental Problems
Solutions and Answers 841
Selected Answers 874
Index 898
717
Chapter Objectives 717
The Elastic Curve 717
Slope and Displacement by
Integration 721
Discontinuity Functions 739
Method of Superposition 750
Statically Indeterminate Beams and
Shafts-Method of Superposition 758
777
Appendix
C
15
XVI I
This page intentionally left blank
STATICS AND
MECHANICS OF
MATERIALS
CHAPTER
1
•- ae-T.
Large cranes such as this one are required to lift extremely large loads. Their
design is based on the basic principles of statics and dynamics, which fonn the
subject matter of engineering mechanics.
GENERAL PRINCIPLES
CHAPTER OBJECTIVES
•
To provide an introduction to the basic quantities and ideal'izations
of mechanics.
•
To state Newton's Laws of Motion.
•
To review the principles for applying the SI system of units.
•
To examine the standard procedures for performing numerical
calculations.
•
To present a general guide for solving problems.
1. 1
MECHANICS
Mechanics can be defined as that branch of the physical sciences concerned
with the state of rest or motion of bodies that are subjected to the action
of forces. In this book we will study two important branches of mechanics,
namely, statics and mechanics of materials. These subjects form a suitable
basis for the design and analysis of many types of structural, mechanical,
or electrical devices encountered in engineering.
Statics deals with the equilibrium of bodies, that is, it is used to
determine the forces acting either external to the body or within it that
are necessary to keep the body in equilibrium. Mechanics of materials
studies the relationships between the external loads and the distribution
of internal forces acting within the body. This subject is also concerned
with finding the deformations of the body, and it provides a study of the
body's stability.
3
4
CHAPTER 1
GENERAL PRINC I PLES
In this book we will first study the principles of statics, since for the
design and analysis of any structural or mechanical element it is first
necessary to determine the forces acting both on and within its various
members. Once these internal forces are determined, the size of the
members, their deflection, and their stability can then be determined using
the fundamentals of mechanics of materials, which will be covered later.
Historical Development. The subject of statics developed very
early in history because its principles can be formulated simply from
measurements of geometry and force. For example, the writings of
Archimedes (287- 212 a.c.) deal with the principle of the lever. Studies of
the pulley and inclined plane are also recorded in ancient writings - at
times when the requirements for engineering were limited primarily to
building construction.
The origin of mechanics of materials dates back to the beginning of the
seventeenth century, when Galileo performed experiments to study the
effects of loads on rods and beams made of various materials. However, at
the beginning of the eighteenth century, experimental methods for testing
materials were vastly improved, and at that time many experimental and
theoretical studies in this subject were undertaken primarily in France, by
such notables as Saint-Venant, Poisson, Lame, and Navier.
Over the years, after many of the fundamental problems of mechanics
of materials had been solved, it became necessary to use advanced
mathematical and computer techniques to solve more complex problems.
As a result, this subject has expanded into other areas of mechanics, such
as the theory of elasticity andl the theory of plasticity. Research in these
fields is ongoing, in order to meet the demands for solving more advanced
problems in engineering.
1.2
FUNDAMENTAL CONCEPTS
Before we begin our study, it is important to understand the definitions
of certain fundamental concepts and principles.
Mass. Mass is a measure of a quantity of matter that is used to compare
the action of one body with that of another. This property provides a
measure of the resistance of matter to a change in velocity.
Force. In general,force is considered as a "push" or "pull" exerted by
one body on another. This in teraction can occur when there is direct
contact between the bodies, such as a person pushing on a wall, or it can
occur through a distance when the bodies are physically separated.
Examples of the latter type include gravitational, electrical, and magnetic
forces. In any case, a force is completely characterized by its magnitude,
direction, and point of application.
1.2
FUNDAM ENTAL CONCEPTS
Particle. A particle has a mass, but a size that can be neglected. For
example, the size of the earth is insignificant compared to the size of its
orbit, and therefore the earth can be modeled as a particle when studying
its orbital motion. When a body is idealized as a particle, the principles of
mechanics reduce to a rather simplified form since the geometry of the
body will not be involved in the analysis of the problem.
R191d Bod A rigid body can be considered as a combination of a
large number of particles in which all the particles remain at a fixed
distance from one another, both before and after applying a load. This
model is important because the material properties of any body that is
assumed to be rigid will not have to be considered when studying the
effects of forces acting on the body. lo most cases the actual
deformations occurring in structures, machines, mechanisms, and the
like are relatively small, and the rigid-body assumption is suitable for
analysis.
Concentrated Force. A concentrated force represents the effect of
a loading which is ass umed to act at a point on a body. We can represent a
load by a concentrated force, provided the area over which the load is
applied is very small compared to the overall size of the body. An example
would be the contact force between a wheel and the ground.
Steel is a common e ngineering material that does not
deform very much under load. Therefore, we can consider
this rai lroad wheel to be a rigid body acted upon by the
concentrated force o f the rail.
Three forces act on the ring. Since these
forces all meet at a point, then for any
force analysis, we can assume the ring to
be represented as a particle.
5
6
CHAPTER 1
GENERAL PRINC I PLES
Newton's Three Laws of Motion. Engineering mechanics is
formulated on the basis of Newton's three Jaws of motion, the validity of
which is based on experimental observation. These Jaws apply to the
motion of a particle as measured from a nonaccelerating reference frame.
They may be briefly stated as follows.
First Law. A particle originally at rest, or moving in a straight line with
constant velocity, tends to remain in this equilibrium state provided the
particle is not subjected to an unbalanced force, Fig. 1- la.
·y ·,.
F1
Equilibrium
(a)
Second Law. A particle acted upon by an unbalanced force F
experiences an acceleration a that has the same direction as the force
and a magnitude that is directly proportional to the force, Fig. 1- lb. If
the particle has a mass m, this Jaw may be expressed mathematically as
(1- 1)
F = ma
a
Accelerated motion
(b)
Third Law. The mutual forces of action and reaction between two
particles are equal, opposite, and collinear, Fig. 1- lc.
torce of A on B
F
0@3
A
B
F
\_force of Bon A
Action - reaction
(c)
Fig.1-1
1.2
FUNDAM ENTAL CONCEPTS
Newton's Law of Gravitational Attraction. Shortly after
formulating his three laws o f motion, Newton postulated a law governing
the gravitational attraction between any two particles.Stated mathematically,
F
=G
1111 ,,,,,
r2
-
(1-2)
where
F= force of gravitation between the two particles
G =universal constant of gravitation; according to experimental
evidence, G = 66.73( 10-12 ) m3 /( kg· s2 )
111., 1112 = mass of each of the two particles
r =distance between the two particles
Weight. According to Eq. l-2, any two particles or bodies have a
mutual attractive (gravitational) force acting between them. In the case
of a particle located at or near the surface of the earth, however, the only
gravitational force having any sizable magnitude is that betwe,e n the
earth and the particle. Consequently, this force, called the weight, will be
the only gravitational force considered in our study of mechanics.
From Eq. 1-2, we can develop an approximate expression for finding the
weight W of a particle having a mass m1 = m. If we assume the earth to be
a nonrotating sphere of constant density and having a mass 1112 = M,, then
if r is the distance between the earth's center and the particle, we have
mM,
W= G -?r
This astronaut"s weight is diminished since
she is far removed from lhe gravitational
field of the earth. (© NikoNomad/
Shutters1ock)
Letting g = GM,/ r2 yields
(1-3)
By comparison with F = ma, we can see that g is the acceleration due
to gravity. Since it de pends on r, the weight of a particle or body is not
an absolute quantity. Instead, its magnitude is determined from where
the measurement was made. For most engineering calculations, however,
g is determined at sea level and at a latitude of 45°, which is considered
the "standard location."
7
8
CHAPTER 1
GENERAL PRINC I PLES
Name
Length
Time
Mass
Force
International
System of Units
SI
meter
second
kilogram
Inewton* I
m
s
kg
N
(kg ~m)
5
•Derived unit.
1. 3
THE INTERNATIONAL SYSTEM OF
UNITS
The four basic quantities-length, time, mass, and force - are not all
independent from one another; in fact, they are related by Newton's
second Jaw of motion, F = ma. Because of this, the units used to measure
these quantities cannot all be selected arbitrarily. The equality F = ma is
maintained only if three of the four units, called base units, are defined
and the fourth unit is then derived from the equation.
For the International System of Units, abbreviated SI after the French
"Systeme International d'Unites," length is in meters (m), time is in
seconds (s), and mass is in kilograms (kg), Table 1- 1. The unit of
force, called a newton (N), is derived from F = ma. Thus, 1 newton is
equal to a force required to give 1 kilogram of mass an acceleration of
1 m/s2 (N = kg · m/s2 ) .
If the weight of a body located at the "standard location" is to be
determined in newtons, then Eq.1- 3 must be applied. Here measurements
give g = 9.806 65 m/s2 ; however, for calculations the valueg = 9.81 m/s2
will be used. Thus,
W = mg
(g = 9.81 m/s2 )
(1-4)
Therefore, a body of mass 1 kg has a weight of9.81N,a2-kg body weighs
19.62 N, and so on, Fig. 1- 2. Perhaps it is easier to remember that a small
apple weighs one newton. Also, by comparison with the U.S. Customary
system of units (FPS),
Fig.1-2
1 pound (lb)
1 foot (ft)
=
=
4.448 N
0.3048 m
Prefixes. When a numerical quantity is either very large or very small,
the units used to define its size may be modified by using a prefix. Some
of the prefixes used in the SI system are shown in Table 1- 2. Each
represents a multiple or sn.ibmultiple of a unit which, if applied
successively, moves the decimal point of a numerical quantity to every
1.3
THE INTERNATIONAL SYSTEM OF UNITS
third place.* For example, 4 000 000 N = 4 000 kN (kilo-newton) =
4 MN (mega-newton), or 0.005 m = 5 mm (milli-meter). Notice that the
SI system does not include the multiple deca (10) or the submultiple
centi (0.01), which form part of the metric system. Except for some
volume and area measurements, the use of these prefixes is generally
avoided in science and engineering.
Exponential Form
Prefix
SI Symbol
109
106
1()-1
giga
mega
kilo
G
M
k
10-3
10-6
10-9
milli
micro
nano
m
Multiple
1000000 000
1000000
1 000
Submulriple
0.001
0.000 001
0.000 000 001
J.L
n
Rules for Use. Here are a few of the important rules that describe the
proper use of the various SI symbols:
• Quantities defined by several units which are multiples of one
another are separated by a dot to avoid confusion with prefix
notation ' as indicated by N = k<>0 · m/s2 = k<>0 · m · s- 2 . Also' m · s
(meter-second), whereas ms (milli-second).
• The exponential power on a unit having a prefix refers to both the
unit and its prefix. For example, µN 2 = (µN) 2 = µN · µN. Likewise,
mm2 represents (mm)2 = mm· mm.
• With the exception of the base unit the kilogram, in general avoid the
use of a prefix in the denominator of composite units. For ex.ample,
do not write N/mm, but rather kN/m; also, m/mg should be written
as Mm/kg.
• When performing calculations, represent the numbers in terms of
their base or derived units by converting all prefixes to powers of 10.
The final result should then be expressed using a single p refix. Also,
after calculation, it is best to keep numerical values between 0.1 and
1000; otherwise, a suitable prefix should be chosen. For example,
(50 kN)(60 nm) = [50(103 ) N][60(10- 9 ) m]
= 3000(10- 6 ) N · m = 3(10- 3) N · m = 3 mN • m
*The kilogram is the only base unit that is defined with a prefix.
9
10
CHAPTER
1
GENERAL PRI NCIPLES
1.4 NUMERICAL CALCULATIONS
Numerical work in engineering practice is most often performed by using
handheld calculators and computers. It is important, however, that the
answers to any problem be reported with justifiable accuracy using
appropriate significant figures. In this section we will discuss these topics
together with some other important aspects involved in all engineering
calculations.
Dimensional Homogeneity. The terms of any equation used to
Computers are often used in engineering for
advanced design and analysis. (© Blaize
Pascall/Alamy)
describe a physical process must be dimensionally homogeneous; that is,
each term must be expressed in the same units. Provided this is the case,
all the terms of an equation can then be combined if numerical values
are substituted for the variables. Consider, for example, the equation
s = vt + ~ at2, where, in SI units, sis the position in meters, m, tis time
in seconds~s, v is velocity in m/s, and a is acceleration in m/s2 . Regardless
of how this equation is evaluated, it maintains its dimensional
homogeneity. In the form stated, each of the three terms is expressed in
meters [ m, (m/i)i, (m/i>)sl ) or solving for a, a = 2s/t2 - 2v/t, the
terms are each expressed in units of m/s2 [m/s2 , m/s2 , (m/s)/s].
Keep in mind that problems in mechanics always involve the solution
of dimensionally homogeneous equations, and so this fact can then be
used as a partial check for algebraic manipulations of an equation.
Significant Figures. The number of significant figures contained in
any number determines the accuracy of the number. For instance, the
number 4981 contains four significant figures. However, if zeros occur at
the end of a whole number, it may be unclear as to how many significant
figures the number represents. For example, 23 400 might have three
(234), four (2340), or five (23 400) significant figures. To avoid these
ambiguities, we will use engineering notation to report a result. This
requires that numbers be rounded off to the appropriate number of
significant digits and then expressed in multiples of (103) , such as (103) ,
(106) , or (10- 9) . For instance, if 23 400 has five significant figures, it is
written as 23.400(103), but if it has only three significant figures, it is
written as 23.4(103).
If zeros occur at the beginning of a number that is Jess than one, then
the zeros are not significant. For example, 0.008 21 has three significant
figures. Using engineering notation, this number is expressed as 8.21(10- 3).
Likewise, 0.000 582 can be expressed as 0.582(10- 3) or 582(10- 6).
1. 5
GENERAL PROCEDURE FOR ANALYSIS
Rounding Off Numbers. Rounding off a number is necessary so
that the accuracy of the result will be the same as that of the problem
data. As a general rule, any numerical figure ending in a number greater
than five is rounded up and a number less than five is not rounded up. The
rules for rounding off numbers are best illustrated by example. Suppose
the number 3.5587 is to be rounded off to three significant figures. B ecause
the fourth digit (8) is greater than 5, the third number is rounded up
to 3.56. Likewise 0.5896 becomes 0.590 and 9.3866 becomes 9.39. If we
round off 1.341 to three significant figures, because the fourth digit (1) is
less than 5, then we get 1.34. Likewise 0.3762 becomes 0.376 and 9.871
becomes 9.87. There is a special case for any number that ends in a 5. As a
general rule, if the digit preceding the 5 is an even number, then this digit
is not rounded up. lf the digit preceding the 5 is an odd number, then it is
rounded up. Fo r example, 75.25 rounded off to three significant digits
becomes 75.2, 0.1275 becomes 0.128, and 0.2555 becomes 0.256.
Calculations. Whe n a sequence of calculations is performed, it is best
to store the intermediate results in the calculator. In other words, do not
round off calculations until expressing the final result. This procedure
maintains precision throughout the series of steps to the final solution. In
this book we will generally round off the answers to three significant
figures since most of the data in engineering mechanics, such as geometry
and loads, may be reliably measured to this accuracy.
1 .5
GENERAL PROCEDURE FOR
ANALYSIS
Attending a lecture, reading this book, and studying the example problems
helps. but the m ost effedive way o f learning the principles of engineering
mechanics is t o solve problems. To be successful at this, it is important to
always present the work in a logical and orderly manner, as suggested by
the following sequence of steps:
• Read the problem carefully and try to correlate the actual physical
situation with the theory studied.
• Tabulate the problem data and draw to a large scale any necessary
diagrams.
• Apply the relevant principles, generally in mathematical form. When
writing any equations, be sure they are dimensionally homogeneous.
• Solve the necessary eq uations, and report the answer with no more
than three s ignificant figures.
• Study the answer with tec hnical judgment and common sense to
determine whether o r not it seems reasonable.
When solving problems, do the work as
neatly as possible. Being neat will
stimulate clear and orderly thinking,
and vice versa.
11
12
CHAPTER
1
GENERAL PRINC I PLES
IMPORTANT POINTS
• A particle has a mass but a size that can be neglected, and a
rigid body does not deform under load.
• A force is considered as a "push" or "pull" of one body on
another.
• Concentrated forces are assumed to act at a point on a body.
• Newton's three Jaws of motion should be memorized.
• Mass is measure of a quantity of matter that does not change
from one location to another. Weight refers to the gravitational
attraction of the earth on a body or quantity of mass. Its magnitude
depends upon the elevation at which the mass is located.
• In the SI system the unit of force, the newton, is a derived unit.
The meter, second, and kilogram are base units.
• Prefixes G, M, k, m, µ,, and n are used to represent large and small
numerical quantities. Their exponential size should be known,
along with the rules for using the SI units.
• Perform numerical calculations with several significant figures,
and then report the final answer to three significant figures.
• A lgebraic manipulations of an equation can be checked in
part by verifying that the equation remains dimensionally
homogeneous.
• Know the rules for rounding off numbers.
1. 5
EXAMPLE
13
GENERAL PROCEDURE FOR ANALYSIS
1.1
Convert 2 km/ h to m/s. H ow many ft /s is this?
SOLUTION
Since 1 km = 1000 m and 1 h = 3600 s, the factors o f conversion are
arranged in the following order, so that a cancellation of the units can
be applied:
m)(
1 l< )
2 km/ h = 2J,;.m(1000
l<
J,;.m
3600 s
2000m
=
= 0.556 m/s
3600s
Ans.
Since 1 ft = 0.3048 m, then
m)(
0 _556 m/s = (0.556
s
= 1.82 ft/s
I ft
o.3048
)
m
Ans.
NOTE: Remember to round off the final answer to three significant
figures.
EXAMPLE
1 .2
Convert 300 lb· ft to appropriate SI units.
SOLUTION
Since 1 lb
= 4.448 N and 1 ft = 0.3048 m, the n we have
300 lb'· f( = ( 4.448 N) (0.3048 m)
llb'
I f(
=
407N·m
Ans.
14
CHAPTER 1
I EXAMPLE
GENERAL PRI N CIPLES
1.3
Evaluate each of the following and express with SI units having an
appropriate prefix: (a) (50 mN)(6 GN), (b) (400 mm)(0.6 MN)2,
(c) 45 MN3/ 900 Gg.
SOLUTION
First convert each number to base units, perform the indicated
operations, then choose an appropriate prefix.
Part (a)
(50 mN)(6 GN)
= (
50( 10- 3 ) N ][ 6( 109 ) N j
=
300( 106 ) N2
=
300( 106 ) w(
=
300 kN2
1
~)(
~)
10 ]>( 10 ]>(
1
Ans.
NOTE: Keep in mind the convention kN2 = ( kN ) 2 = 106 N2 .
Part (b)
(400 mm)(0.6 MN) 2
=
m ] [ 0.6( 106 ) N ]2
( 400( 10- 3 )
m ][ 0.36( 10 12 ) N2 j
= ( 400( 10- 3)
=
E44( 109 ) m·N2
=
144Gm · N2
Ans.
We can also write
144( 109 ) m · N2
=
144( 109 ) m · w(
=
0.144 m • MN2
1
1
~)(
~N)
10 M 10 N
Ans.
Part (c)
45( 106 N) 3
45 MN3
---900 Gg
900( 106 ) kg
=
50( 109 ) N3/ kg
=
50( 109)
=
w-( 101
50 kN3/ kg
3
kN
)
3
]>(
J...
kg
Ans.
PROBLEMS
15
PROBLEMS
The answers to all but every fourth problem (asterisk) an given in the back of the book.
1-1. What is the weight in newtons of an object that
bas a mass of (a) 8 kg. {b) 0.04 kg, (c) 760 Mg?
1-2. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) kN/ µs. {b) Mg/mN, (e) MN/{kg ·ms).
1-3. Represent each of 1he following combinations of
units in the correct SI form: (a) Mg/ms, (b) N/mm,
(c) mN/(kg · µs).
*1-4. Convert: (a) 200 lb· fl to N · m, (b) 350 lb/ft3 to kN/m3,
(c) 8 ft/h to mm/s. Express the resu lt to three significant
figures. Use an appropriate prefix.
1-5. Represent each of the following as a number between
0.1 and 1000 using an appropriate prefix: (a) 45320 kN,
(b) 568(105) mm,(e) 0.00563 mg.
1-6. Round off lhc following numbers to three significant
figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, (d) 7555 kg.
1-7. Represent each of the following quantities in the
correct SI form using an appropriate prefix: (a) 0.000 431 kg,
(b) 35.3( Iol) N. (c) 0.005 32 km.
*1-8. Represent each of the following combinations of units
in the correct SI form using an appropriate prefix: (a) Mg/mm,
(b) mN/µs.(c) µm ·Mg.
1-9. Represent each of the following combinations of
units in the correct SI form using an appropriate prefix:
(a) m/ms, (b) µkm. (c) ks/mg, (d) km· µN.
1-10. Represent each of 1he following combinations of
units in the correct SI form using an appropriate prefix:
(a) GN · µm, (b} kg/µm, (c) N/ks2, and {d) kN/µs.
1-11. Represent each of the following with SI units having
an appropriate prefix: (a) 8653 ms, (b} 8368 N, (c) 0.893 kg.
*1-U. Evaluate each of the following to three significant
figures and express each answer in SI units using
an appropriate prefix: (a)
(684 µm)/(43 ms),
(b) (28 ms)(0.0458 Mm)/(348 mg), (c) (2.68 mm)(426 Mg).
1-13. Coovert each of the following to three significant
figures. (a) 20 lb· ft to N · m. {b) 450 lb/ft3 to kN/m3 ,
(c) 15 ft/b to mm/s.
1-14. Evaluate each of the following to three significant
figures and express each answer in SI units using an
appropriate prefix: (a) {212 mN)2, {b) {52 800 ms)2,
(c) [548(106)] 112 ms.
1-15. Using the SI system of units, show that Eq. 1- 2 is a
dimensionally homogeneous equation which gives F in
newtons. Determine to three significant figures the
gravitational force acting between two spheres that are
touching each other. The mass of each sphere is 200 kg and
the radius is 300 mm.
*1-16. The pascal (Pa) is actually a very small unit of
pressur·e. To show this, convert 1 Pa = l N/m2 to lb/ft2 .
Atmosphere pressure al sea level is 14.7 lb/in 2. How many
pascals is this?
1-17. What is the weight in newtons of an object that has
a mass of: (a) 10 kg. (b) 0.5 g. (c) 4.50 Mg? Express the
result to three significant figures. Use an appropriate prefix.
1-18. Evaluate each of the following to three significant
figures and express each answer in SI units using an
appropriate prefix: (a) 354 mg(45 km)/(0.0356 kN),
(b) (0.004 53 Mg)(20l ms). (c) 435 MN/23.2 mm.
1-19. A concrete column has a diameter of 350 mm and
a length of2 m. Uthe density (mass/volume) of concrete is
2.45 Mg/ml, determine the weight of the column in pounds.
*1-20. Two particles have a mass of 8 kg and 12 kg,
respectively. U they are 800 mm apart, determine the force
of gravity acting between them. Compare this result with
the weight of each particle.
1-2L U a man weighs 155 lb o n earth, specify (a) his
mass in kilograms, and (b) his weight in newtons. If the
man is on the moon , where the acce leration due to gravity
is g., = 5.30 ft/s 2, determ ine (c) his weight in pounds, and
(d) his mass in kilograms.
CHAPTER
/
,,,,I'
,~
,'/
'
'1
}
·1
1
/
//7
II
1'
"
I
1
~
\
\I
I
\
I
\
!
,
I
\\
\
\
(© Vasiliy Koval/Fotolia)
This electric transmission tower is stabilized by cables that exert forces on the
tower at their points of connection. In this chapter we will show how to express
these forces as Cartesian vectors, and then determine their resultant.
FORCE VECTORS
CHAPTER OBJECTIVES
•
To show how to add forces and resolve them into components
using the Parallelogram Law.
•
To express force and position as Cartesian vectors.
•
To introduce the dot product in order to use it to find the angle
b etween two vectors or the projection of one vector onto another.
2.1
SCALARS AND VECTORS
Many physical quantities in engineering mechanics are measured using
either scalars or vectors.
Scalar. A scalar is any positive or negative physical quantity that can
be completely specified by its magnitude. Examples of scalar quantities
include length, mass, and time.
Vector. A vector is any physical quantity that requires both a magnitude
and a direction for its complete description. Examples of vectors
encountered in statics are force, position, and moment. A vector A
is shown graphically by an arrow, Fig. 2- 1. The length of the arrow
represents the magnitude of the vector, and the angle 8 between the vector
and a fixed axis defines the direction of its line of action.The head or tip of
the arrow indicates the sense of direction of the vector.
In print, vector quantities are represented by boldface letters such as
A , and the magnitude of a vector is italicized, A. For handwritten work, it
is often convenient to denote a vector quantity by simply drawing an
. ~
arrow above 1t, A.
Line of action~ _
~
Tail~
0
Fig. 2-1
17
18
CHAPTER
2
FORCE VECTORS
2.2
VECTOR OPERATIONS
Multiplication and Division of a Vector by a Scalar. If a
vector is multiplied or divided by a positive scalar, its magnitude is
increased by that amount. Multiplying or dividing by a negative scalar
will also change the directional sense of the vector. Graphic examples of
these operations are shown in Fig. 2- 2.
Vector Addition. When adding two vectors together it is important
Scalar multiplication and division
Fig. 2-2
to account for both their magnitudes and their directions. To do this we
must use the parallelogram law. To illustrate, the two component vectors
A and B in Fig. 2- 3a are addled to form a resultant vector R = A + B
using the following procedure:
•
First join the tails of the components at a point to make them
concurrent, Fig. 2- 3b.
•
From the head of B, draw a line parallel to A. Draw another line
from the head of A that is parallel to B. These two lines intersect at
point P to form the adjacent sides of a parallelogram.
The diagonal of this parallelogram that extends to P forms R , which
then represents the resultant vector R = A + B, Fig. 2- 3c.
•
p
R =A+ B
(a)
Parallelogram law
(c)
(b)
Fig. 2-3
We can also add B to A , Fig. 2-4a, using the triangle rule, which is a
special case of the parallelogram law, whereby vector B is added to
vector A in a "head-to-tail" fashion , i.e., by connecting the tail of B to
the head of A , Fig. 2-4b. The resultant R extends from the tail of A to
the head of B. In a similar manner, R can also be obtained by adding
A to B, Fig. 2- 4c. By comparison, it is seen that vector addition is
commutative; in other words, the vectors can be added in either order,
i.e., R = A + B = B + A.
2.2
R
R=A+B
R = B+A
Triangle rule
Triangle rule
(b)
(c)
(a)
Fig. 2-4
As a special case, if the two vectors A and B are collinear, i.e., both
have the same line of action, the parallelogram Jaw reduces to an
algebraic or scalar addition R = A + B, as shown in Fig. 2- 5.
R
•
•
A
B
R=A+B
Addition of collinear vectors
Fig. 2-5
Vector Subtraction. The resultant of the difference between two
vectors A and B of the same type may be expressed as
R' =A - B =A + (-B)
This vector sum is shown graphically in Fig. 2-6. Subtraction is therefore
defined as a special case of addition, so the rules of vector addition also
apply to vector subtraction.
-B
or
B
-B
R'= A-B
Parallelogram Jaw
Vector s ubtraction
Fig. 2-6
R' = A-B
Triangle rule
VECTOR OPERATIONS
19
20
CHAPTER
2
FORCE VECTORS
2. 3
VECTOR ADDITION OF FORCES
Experimental evidence has shown that a force is a vector quantity since
it has a specified magnitude, direction, and sense and it adds according to
the parallelogram Jaw. Two common problems in statics involve either
finding the resultant force, knowing its components, or resolving a known
force into two components. We will now describe how each of these
problems is solved using the parallelogram Jaw.
Finding a Resultant Force. The two component forces F 1 and F2
acting on the pin in Fig. 2- 7a are added together to form the resultant
force FR = F 1 + F2, using the parallelogram Jaw as shown in Fig. 2- 7b.
From this construction, or using the triangle rule, Fig. 2- 7c, we can then
apply the Jaw of cosines or the Jaw of sines to the triangle in order to
obtain the magnitude of the resultant force and its direction.
The parallelogram law must be used to
determine the resultant of the two
forces acting on the hook.
F
(a)
II
(b)
(c)
Fi.g. 2-7
Finding the Components of a Force. Sometimes it is necessary
to resolve a force into two components in order to study its pulling or
'
Using the parallelogram law the
supporting force F can be resolved into
components acting along the u. and v axes.
pushing effect in two specific directions. For example, in Fig. 2-8a, F is to
be resolved into two components along the two members, defined by the
u and v axes. In order to determine the magnitude of each component, a
parallelogram is constructed first, by drawing lines starting from the tip of
F, one line parallel to u, and the other line parallel to v. These lines
intersect with the v and u axes, forming a parallelogram. The force
components F11 and Fv are established by simply joining the tail of F to the
intersection points on the u and v axes, Fig. 2-8b. This parallelogram can
be reduced to a triangle, which represents the triangle rule, Fig. 2- &. From
this, the Jaw of sines can be applied to determine the unknown magnitudes
of the components.
2.3
VECTOR ADDITION OF FORCES
21
v
F,.
F,,
(b)
(a)
(c)
Fig. 2-8
Addition of Several Forces. If more than two forces are to be
added, successive applications of the parallelogram law can be carried
out in order to obtain the resultant force. For example, if three forces Fi,
F2, F 3 act at a point 0 , Fig. 2- 9, the resultant of any two of the forces is
found, say, F1 + F2, and then this resultant is added to the third force,
yielding the resultant of all three forces; i.e., FR= (F1 + F2) + F3 • Using
the parallelogram law to add more than two forces, as shown here,
generally requires extensive geometric and trigonometric calculation to
determine the magnitude and direction of the resultant. Instead,
problems of this type are easily solved by using the "rectangularcomponent method," which is explained in the next section.
Fig. 2-9
The resultant force FR on the hook requires
the addition of F1 + F2 , then this resultant is
added to F3 .
22
CHAPTER
2
FORCE VECTORS
IMPORTANT POINTS
• A scalar is a positive or negative number.
• A vector is a quantity that has a magnitude, direction, and sense.
• Multiplication or division of a vector by a scalar will change the
magnitude of the vector. The sense of the vector will change if the
scalar is negative.
• Vectors are added or subtracted using the parallelogram Jaw or
the triangle rule.
• As a special case, if the vectors are collinear, the resultant is
formed by an algebraic or scalar addition.
PROCEDURE FOR ANALYSIS
Problems that involve the addition of two forces can be solved as
follows:
(a)
Parallelogram Law.
• Sketch the addition of the two "component" forces F 1 and F2
according to the parallelogram Jaw, yielding the resultant force
FR that forms the diagonal of the parallelogram, Fig. 2- lOa.
__.-II
A
(b)
• If a force F is to be resolved into components along two axes u
and v, then start at the head of force F and construct lines parallel
to the axes, thereby forming the parallelogram, Fig. 2- lOb. The
sides of the parallelogram represent the components, F11 and Fv.
c
• Label all the known and unknown force magnitudes and the angles
on the sketch and identify the two unknowns as the magnitude
and direction of FR• or the magnitudes of its components.
B
a
b
Trigonometry.
c
Cosine law:
C = -.JA 2 + 8 2 - 2AB cos c
Sine law:
_A_ = _J}_ = _s;_
sin a sin b sin c
(c)
Fig. 2-10
• Redraw a half portion of the parallelogram to illustrate the
triangular head-to-tail addition of the components.
• From this triangle, the magnitude of the resultant force can be
determined using the Jaw of cosines, and its direction is
determined from the Jaw of sines. The magnitudes of two force
components are determined from the Jaw of sines. The formulas
are given in Fig. 2- lOc.
2.3
-
EXAMPLE
23
VECTOR ADDITION OF FORCES
2.1
-
The screw eye in Fig. 2- lla is subjected to two forces, F1 and F2.
Determine the magnitude and direction of the resultant force.
A
65°
l(f
360" - 2(65°)
v - - - - - =us•
2
90• - 25• = 65°
(a)
(b)
SOLUTION
Parallelogram Law. The parallelogram is formed by drawing a line
from the head of F 1 that is parallel to F 2, and another line from the head
of F2 that is parallel to F 1. The resultant force FR extends to where these
lines intersect at point A, Fig. 2- llb. The two unknowns are the
magnitude of FR and the angle 8 (theta).
Trigonometry. From the parallelogram, the vector triangle is shown in
Fig. 2- llc. Using the Jaw of cosines
FR = Y(l00N)2
=
Y 10 000
=
213 N
+ (150N)2
-
2(100N)(l50N)cos 115°
+ 22 500 - 30 000(-0.4226)
=
212.6 N
Ans.
Applying the Jaw of sines to determine 8,
150 N
sin 8
212.6 N
sin 115°
Fig. 2-11
150 N .
0
. N (sm 115 )
212 6
8 = 39.8°
sin 8 =
Thus, the direction </> (phi) of FR, measured from the horizontal, is
</> = 39.8°
+ 15.0°
=
54.8°
Ans.
FR to have
a magnitude larger than its components and a direction that is
between them.
NOTE: The results seem reasonable, since Fig. 2- llb shows
24
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.2
Resolve the horizontal 600-Jb force in Fig. 2- 12a into components
acting along the u and v axes and determine the magnitudes of these
components.
II
600 Jb
600 Jb
c
v
I
v
(a)
(c)
(b)
Fig. 2-U
SOLUTION
Parallelogram Law. The parallelogram is constructed by extending a
line from the head of the 600-Jb force parallel to the v axis until it intersects
the u axis at point B, Fig. 2- 12b. The arrow from A to B represents F11 •
Similarly, the line extended from the head of the 600-Jb force drawn
parallel to the u axis intersects ~he v axis at point C, which gives Fv.
The vector addition using the triangle rule is shown
in Fig. 2- 12c. The two unknowns are the magnitudes of F 11 and Fv.
Applying the Jaw of sines,
Trigonometry.
F,,
sin 120°
600 lb
sin 30°
F,,
= 1039 lb
Fv
600 lb
sin 30°
sin 30°
Fv
= 600 lb
Ans.
Ans.
NOTE: The result for F11 shows that sometimes a component can have
a greater magnitude than the resultant.
2.3
-
EXAMPLE
VECTOR ADDITION OF FORCES
2.3
-
Determine the magnitude of the component force Fin Fig. 2- 13a and
the magnitude of the resultant force FR if FR is directed along the
positive y axis.
y
y
I
/
45•
FR
(a)
FR
(c)
(b}
Fig. 2-13
SOLUTION
The parallelogram Jaw of addition is shown in Fig. 2- 13b, and the
triangle rule is shown in Fig. 2- 13c. The magnitudes of FR and Fare the
two unknowns. They can be determined by applying the Jaw of sines.
F
sin 60°
200 lb
sin 45°
F = 245lb
FR
sin 75°
Ans.
200 lb
sin 45°
FR = 273 lb
Ans.
It is strongly suggested that you test yourself on the solutions to these
examples by covering them over and then trying to draw the
parallelogram law, and thinking about how the sine and cosine laws
are used to determine the unknowns. Then before solving any of the
problems, try to solve the Preliminary Problems and some of the
Fundamental Problems given on the next pages. The solutions and
answers to these are given in the back of the book. Doing this
throughout the book will help immensely in developing your problemsolving skills.
25
26
CHAPTER
2
FORCE VECTORS
PRELIMINARY PROBLEMS
Partial solutions and answers to all Preliminary Problems are given in the back of the book.
P2-L In each case, construct the parallelogram law to
show FR= F 1 + F2 • Then establish the triangle rule, where
FR= F 1 + F2• Label all known and unknown sides and
internal angles.
P2-2. In each case, show how to resolve the force F into
components acting along the u and v axes using the
parallelogram law. Then establish the triangle rule to show
FR = F,. + Fv. Label all known and unknown sides and
interior ang,les.
F= 200N
v
F 1 =200N
II
/
/\45°
(a)
(a)
F= 400N
v
(b)
II
(b)
F=600N
ll
(c)
Prob. P2-1
(c)
Prob. P2-2
2.3
VECTOR ADDITION OF FORCES
27
FUNDAMENTAL PROBLEMS
Partial solutions and answers to all Fundamental Problems are given in the back of the book.
F2-1. Determine the magnitude of the resultant force
acting on the screw eye and its direction measured clockwise
from the x axis.
F2-4. Resolve the 30·1b force into components along the
u and v axes, and determine the magnitude of each of these
components.
v
Prob.F2-1
Prob.F2-4
F2-2. Determine the magnitude of the resultant force.
F2-5. Resolve the force into components acting along
members AB and AC, and determine the magnitude of each
component.
3Cf
I
/
A
40°
200N
Prob.F2-2
F2-3. Determine the magnitude of the resultant force and
its direction measured counterclockwise from the positive
x axis.
y
Prob.F2-5
F2-6. If force Fis to have a component along the u axis of
Fu= 6 kN, determine the magnitude of F and the magnitude
of its component Fv along the v axis.
"
I
800N
F
Prob.F2-3
Prob.F2-6
28
CHAPTER
2
FORCE VECTORS
PROBLEMS
2-1. If e = 60° and F = 450 N, determine the magnitude
of the resultant force and its direction, measured
counterclockwise from the positive x axis.
*2-4. Determine the magnitudes of the two components
of F directed along members AB and AC. Set F = 500 N.
2- 5. Solve Prob. 2-4 with F = 350 lb.
2-2. If the magnitude of the resultant force is to be 500 N,
directed along the positive y axis, determine the magnitude
of force F and its direction 8.
)'
A
700N
c
Probs. 2-112
Probs. 2-4/5
2-3. Determine the magnitude of the resultant force
FR = F1 + Fi and its direction, measured counterclockwise
from the positive x axis.
2-6. Determine the magnitude of the resultant force
FR = F1 + Fi and its direction, measured clockwise from
the positive u axis.
2- 7. Resolve the force F 1 into components acting along
the u and v axes and determine the magnitudes of the
components.
)'
F 1 = 250 lb
*2-8. Resolve the force Fi into components acting along
the u and v axes and determine the magnitudes of the
components.
75°
II
F2 = 375 lb
Prob. 2-3
F2 =6kN
Probs. 2-617/8
2.3
2-9. If the resultant force acting on the suppon is to be
1200 lb, directed horizontally to the right. determine the
force F in rope A and the corresponding angle 8.
V ECTOR ADDITION OF FORCES
29
2-13. The force acting on the gear tooth is F = 20 lb.
Resolve this force into two components acting along the
lines aa and bb.
2-14. The component of force F acting along line aa is
required to be 30 lb. Determine the magnitude of F and its
component along line bb.
-60"
9001b
b
F
a
Prob.2-9
2-10. Determine the magnitude of the resultant force and its
direction. measured counterclockwise from the positive x axis.
y
Probs. 2-13/14
5001b
Prob. 2-10
2-11. If 8 = 60°, determine the magnitude of the resultant
and its direction measured clockwise from the horizontal.
*2-12. Determine the angle 8 for connecting member A
to the plate so that the resultant force of FA and F8 is
directed horizontally to the right. Also. what is the
magnitude of the resultant force?
Probs. 2-lli12
2-15. Force F acts on the frame such that its component
acting along member AB is 650 lb, directed from B
towards A, and the component acting alo11g member BC is
500 lb, directed from B towards C. Determine the magnitude
of F and its direction 8. Set 4> = 6Cl°.
*2-16. Force F acts on the frame such that its component
acting along member AB is 650 lb, directed from B
towards A. Determine the required angle </J (0° s 4> < 45°)
and the component acting along member BC.Set F = 850 lb
and 8 = 300.
Probs. 2-15/16
30
2
CHAPTER
FORCE VECTORS
2-17. If F 1 = 30 lb and F2 = 40 lb, determine the angles IJ
and </> so that the resultant force is directed along the
positive x axis and has a magnitude of FR= 60 lb.
*2-20. Determine the magnitude of force F so that the
resultant FR of the three forces is as small as possible. What
is the minimum magnitude of FR?
8kN
)'
Prob. 2-20
Prob. 2-17
2-18. Determine the magnitude and direction IJ of FA so
that the resultant force is directed along the positive x axis
and has a magnitude of 1250 N.
2-19. Determine the magnitude of the resultant force acting
on the ring at 0 if FA = 750 N and IJ = 45°. What is its
direction, measured counterclockwise from the positive x axis?
2-21. If the resultant force of the two tugboats is 3 kN,
directed along the positive x axis, determine the required
magnitude of force F8 and its direction IJ.
2-22. If F8 = 3 kN and IJ = 45°, determine the magnitude
of the resultant force and its direction measured clockwise
from the positive x axis.
2-23. If the resultant force of the two tugboats is required
to be directed towards the positive x axis, and F8 is to be a
minimum, d etermine the magnitude of FR and F8 and the
angle IJ.
y
A
c
B
Probs. 2-18/19
Probs. 2-21122/23
2.4
2.4
31
ADDITION OF A SYSTEM OF COPLANAR FORCES
ADDITION OF A SYSTEM OF
COPLANAR FORCES
When a force is resolved into two components along the x and y axes, the
components are then called rectangular w mponents. For analytical work
we can represent these components in one of two ways, using either scalar
notation or Cartesian vector notation.
Scalar Nota
The recrangular components of force F shown in
Fig. 2-14a are found using the parallelogram law, so that F = Fx + F,..
Because these components form a right triangle, they can be determined
from
Fx = F cos (}
and
F,, = F sin (}
1
y
Instead of using the angle(}, however, the direction ofF can also be defined
using a small "slope" triangle, as in the example shown in Fig. 2-15b.
Since this triangle and the larger shaded triangle are similar, the proportional
length of the sides gives
F,
(a)
Fx
a
-=F
c
y
or
F= F(a)
C
.T
and
F,,
b
-= -
F
c
or
Here they component is a negative scalar since F,, is directed along the
negative y axis.
It is important to keep in mind that this positive and negative scalar
notation is to be used only for calculations, not for graphical
representations in figures. Throughout the book, the head of a vector
arrow in any figure indicates the sense of the vector graphically; algebraic
signs are not used for this purpose. Thus, the vectors in Figs. 2-14a and
2-14b are designated by using boldface (vector) notation.* Whenever
italic symbols are written near vector arrows in figures, they indicate the
magnitude of the vector, which is always a positive quantity.
• Negative signs are used on ly in figures with boldface notation when showing equal but
opposite pairs of vectors. as in fig. 2-2.
F
(b)
Fig. 2-14
32
CHAPTER
2
FORCE VECTORS
y
Cartesian Vector Notation. Rather than representing the
magnitude and direction of the components Fx and Fy as positive or
negative scalars, we can instead consider them to be only positive scalars
and thereby only report the magnitudes of the components. Their
directions are then represented by the Cartesian unit vectors i and j ,
Fig. 2- 15. These are called unit vectors because they have a dimensionless
magnitude of 1. By separating the magnitude and direction of each
component, we can express Fas a Cartesian vector.
jt
T;------=F
Fy
1"---~---X
____,.....
I
Fig. 2-15
y
Coplanar Force Resultants. We can use either of the two methods
just described to determine the resultant of several coplanar forces, i.e.,
forces that all lie in the same plane. To do this, each force is first resolved
into its x and y components, and then the respective components are added
using scalar algebra since they are collinear. The resultant force is then
formed by adding the resultant components using the parallelogram law.
For example, consider the three concurrent forces in Fig. 2- 16a, which have
x and y components shown in Fig. 2- 16b. Using Cartesian vector notation,
each force is first represented as a Cartesian vector, i.e.,
(a)
y
F1 = F 1xi + F1yj
F2 = -F2x i + F2y j
F3 = F3xi - F3yj
x
The vector resultant, Fig. 2- 17c, is therefore
(b)
FR = F, + F2 + F3
= Fixi + F, y j - F2x i + F2y j + F3x i - F3y j
= (Fi x - F2x + F3x) i + (F1 y + F2y - F3y) j
= (FR)x i + (FR)y j
Fig. 2-16
If scalar notation is used, then indicating the positive directions of
components along the x and y axes with symbolic arrows, we have
...±.
The resultant force of the four cable forces
acting on the post can be dete rmined by
adding algebraically the separate x and y
compone nts of e ach cab le force. This
resultant FRproduces the same pulling effect
on the post as all four cables.
+j
(FR)x = Fix - F2x + F3x
(FR)y = F1y + F1y - F3y
Notice that these are the same results as the i and j components of FR
determined above.
2.4
ADDITION OF A SYSTEM OF COPLANAR FORCES
In general, then, the components of the resultant force of any number
of coplanar forces can be represented by the algebraic sum of the x and y
y
components of all the forces, i.e.,
(FR)x = lFx
(2- 1)
(FR)y = lFy
Once these components are determined, they may be sketched along
the x and y axes with their proper sense of direction, and the resultant
force can be determined from vector addition, Fig. 2- 16c. From this
sketch, the magnitude ofFR is then found from the Pythagorean theorem;
that is,
Also, the angle 8, which specifies the direction of the resultant force, is
determined from trigonometry.
The above concepts are illustrated numerically
which follow.
in
the examples
IMPORTANT POINTS
• The resultant of several coplanar forces can easily be
determined if an x, y coordinate system is established and the
forces are resolved into components along the axes.
• The direction of each force is specified by the angle its line of
action makes with one of the axes, or by a slope triangle.
• The orientation of the x and y axes is arbitrary, and their
positive direction can be specified by the Cartesian unit vectors
i and j.
• The x and y components of the resultant force are simply the
algebraic addition of the components of all the coplanar forces.
• The magnitude of the resultant force is determined from the
Pythagorean theorem, and when the resultant components are
sketched on the x and y axes, Fig. 2- 16c, the direction() of the
resultant can be determined from trigonometry.
(c)
Fig. 2-16 (cont.)
33
34
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.4
Determine the x and y components of F1 and F2 acting on the boom
shown in Fig. 2- 17a. Express each force as a Cartesian vector.
Y
SOLUTION
Scalar Notation. By the parallelogram Jaw, F1 is resolved into x and y
components, Fig. 2- 17b. Since F 1, acts in the - x direction, and F 1>' acts in
the +y direction, we have
F1x = -200 sin 30° N = -100 N = 100 N ~
Ans.
F1y = 200 cos 30° N = 173 N = 173 N f
Ans.
The force F2 is resolved into its x and y components, as shown in
Fig. 2-17c. From the "slope triangle" we could obtain the angle 8,
e.g.,() = tan- 1( 152), and then proceed to determine the magnitudes of
the components in the same manner as for F 1. The easier method,
however, consists of using proportional parts of similar triangles, i.e.,
(a)
y
F 1 = 200 ~
N--.
'\
.F 1y = 200 cos 3D° N
\
260 N
\
\ 30°
\.~-
F
2x
= 260 N(.!3.)
13 = 240 N
Similarly,
\
__
13
\
,.,_~_.._----~ X
Fix = 200 sin 3D° N
F2 = 260 ,../
(b)
>'
y
....
Fix= 260
G~) N
2-)
"\. 13
=
100 N
Notice how the magnitude of the horizontal component, F2x , was
obtained by multiplying the force magnitude by the ratio of the
horizontal leg of the slope triangle divided by the hypotenuse; whereas
the magnitude of the vertical component, F2..,, was obtained by
multiplying the force magnitude by the ratio of the vertical leg divided
by the hypotenuse. Using scalar notation to represent the components,
we have
----p.,.~~~~~~ x
( 5)
s1'!.3.
12
.......
F2y = 260 13 N J-'------'
'~,.,
Fi=260N
(c)
Fig. 2-17
= 240 N = 240 N ~
Ans.
F?..y = -lOON = IOON!
Ans.
F2x
Cartesian Vector Notation. Having determined the magnitudes
and directions of the components of each force, we can express each
force as a Cartesian vector.
F 1 = {-lOOi +173j }N
Ans.
F2 = {240i - lOOj }N
Ans.
2.4
EXAMPLE
ADDITION OF A SYSTEM OF COPLANAR FORCES
2 .5
The link in Fig. 2-18a is subjected to two forces F 1 and F2. D etermine the
magnitude and direction of the resultant force.
y
SOLUTION I
Scalar Notation. Ftrst we resolve each force into its x and y
components, Fig. 2-18b, then we sum these components algebraically.
+ (FR)x
--'-+
= "i.Fx;
(FR)x
= 600 cos 30° N -
400 sin 45° N
= 236.8 N --'-+
(FR)y
(a)
= 600 sin 30° N + 400 cos 45° N
= 582.8 Nf
The resultant force, shown in Fig. 2- 18c, has a magnitude of
FR = V(236.8 N)2
y
+ (582.8 N)2
= 629N
Ans.
From the vector addition,
_
_ (582.8 N) _
- 67.90
8 - tan 1
23.
68N
Ans.
(b)
SOLUTION II
Cartesian Vector Notation. From Fig. 2- 18b, each force is first
expressed as a Cartesian vector.
F1 = { 600 cos 30°i + 600 sin 30°j } N
F2 = {-400 sin 45°i + 400 cos 45°j } N
Then,
y
582.SN
FR = F1
,..1----,, FR
+ F2 = (600 cos 300 N - 400 sin 45° N)i
+ (600 sin 30° N + 400 cos 45° N)j
= { 236.8i + 582.8j } N
The magnitude and direction of FR are determined m the same
manner as before.
(c)
Fig. 2-18
NOTE: Comparing the two m ethods of solution, notice that the use
of scalar nota tion is more efficient since the components can be
found directly, without first having to express each force as a
Cartesia n vector before adding the components. Later, however, we
will show that Cartesian vector analysis is very beneficial for solving
three-dimensional problems.
35
36
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.6
The end of the boom 0 in Fig. 2- l9a is subjected to the three concurrent
and coplanar forces. Determine the magnitude and direction of the
resultant force.
y
(a)
SOLUTION
Each force is resolved into its x and y components, Fig. 2- 19b. Summing
the x components, we have
y
~ (FR)x
= lFx;
(FR)x =
=
-400 N + 250 sin 45° N - 200 ( ~) N
-383.2 N
=
383.2 N ~
Summing they components yields
+ f (FR)y =
lFy;
(FR)y =
(b}
=
250 cos 45° N + 200 ( ~) N
296.8 Nf
The resultant force, shown in Fig. 2- 19c, has a magnitude of
FR =
y
I
296.8 N
=
V(-383.2 N)2 + (296.8 N)2
485N
Ans.
From the vector addition in Fig. 2- 19c, the direction angle() is
8 = tan- 1(296.8) = 37 .80
0
383.2 N
383.2
Ans.
NOTE: Application of this method is more convenient, compared to
(c)
Fig. 2-19
using two applications of the parallelogram Jaw, first to add F 1 and F2,
then adding F3 to this resultant.
2.4
ADDITION OF A SYSTEM OF COPLANAR FORCES
37
FUNDAMENTAL PROBLEMS
Fl-7. Resolve each force into its x and y components.
y
IF1
= 300N
• 2-1 '· If the resultant force acting on the bracket is to be
750 N directed along the positive x axis, determine the
magnitude of F and its direction 8.
y
325 N
Prob.F2-7
F2-8. Determine the magnitude and direction of the
resultant force.
y
250N
400 N
Prob.F2-10
F2-11. If the magnitude of the resultant force acting on
the bracket is to be 80 lb directed along the 11 axis, determine
the magnitude of F and its direction 8.
y
.... .• .... ·.·
••
p"
.l-8
.·2-···
Determine the magnitude of the resultant force
acting on the corbel and its direction 8. measured
counterclockwise from the x axis.
y
F3 = 600 lb
Pr•• rl-t:
• '2-12. Determine the magnitude of the resultant force
and its direction 8. measured counterclockwise from the
positive x axis.
F2 = 400 lb
F1 =15 kN
Prob. F2-9
Prob. F2-12
38
CHAPTER
2
FORCE VECTORS
PROBLEMS
*2-24. Determine the magnitude of the resultant force
and its direction, measured counterclockwise from the
positive x axis.
2- 26. Resolve F1 and F2 into their x and y components.
2- 27. Determine the magnitude of the resultant force and
its direction measured counterclockwise from the positive x
axis.
)'
F2 = 150 N
45°
.../
x
Prob. 2-24
Probs. 2-26/27
2-25. Determine the magnitude of the resultant force and
its direction, measured clockwise from the positive x axis.
*2-28. Resolve each force acting on the gusset plate into
its x and y components, and express each force as a
Cartesian vector.
2- 29. Determine the magnitude of the resultant force
acting on the gusset plate and its direction, measured
counterclockwise from the positive x axis.
)'
400N
)'
/ - -2
~
800N
Prob. 2-25
Probs. 2-28/29
=750N
2.4
2-30. Express each of the three forces acting on the
support in Cartesian vector form and determine the
magnitude of lhe resultanl force and its direction, measured
clockwise from posilive x axis.
ADDITION OF A SYSTEM OF COPLANAR FORCES
2-34.
Express F1, F2, and F3 as Cartesian vectors.
2-35. Determine the magnilude of the resultant force and its
direction, measured counlerclockwise from the positive x axis.
)'
F1 =50N
)'
39
F1=850N
Probs. 2-34135
Prob. 2-30
2-31. Delermine lhe x and y components of F 1 and F2 •
*2-32. Determin e lhe magnilude of the resultant force
and its direction, measured counterclockwise from the
positive x axis.
*2-36. Determine the magnitude of the resultant force
and its direction, measured clockwise from the positive
x axis.
)'
40 1b
)'
Probs. 2-31132
2-33. Determine the magnitude of the resultant force and
its direction. measured counterclockwise from the positive x
axis.
91 lb
)'
Prob. 2-36
F2 = 5 kN
2-37. Determine the magnitude and direction 8 of the
resultant force FR. Express the result in terms of the
magnitudes of the components F1 and F2 and the angle c/J.
Fz
Prob. 2-33
Prob. 2-37
40
CHAPTER
2
FORCE VECTORS
2. 5
y
x
CARTESIAN VECTORS
The operations of vector algebra, when applied to solving problems in
three dimensions, are greatly simplified if the vectors are first represented
in Cartesian vector form. In this section we will present a general method
for doing this; then in the next section we will use this method for finding
the resultant force of a system of concurrent forces.
Right-Handed Coordinate System. We will use a right-handed
Fig. 2-20
coordinate system to describe the vector algebra that follows.
Specifically, a rectangular coordinate system is said to be right handed if
the thumb of the right hand points in the direction of the positive z axis
when the right-hand fingers are curled about this axis and directed from
the positive x towards the positive y axis, Fig. 2- 20.
Rectangular Components of a Vector. In general a vector A
will have three rectangular components along the x, y, z coordinate axes,
Fig. 2- 21. These components are determined using two successive
applications of the parallelogram Jaw, that is, A = A' + Az and then
A ' = Ax + Ar Combining these equations to eliminate A', A 1s
represented by the vector sum of its three rectangular components,
z
I
A,
(2- 2)
Cartesian Vector Representation. In three dimensions, the set
of Cartesian unit vectors, i, j , k, is used to designate the directions of the
x, y, z axes, respectively, Fig. 2- 22. Using these vectors, the three
components of A in Fig. 2- 23 can be written in Cartesian vector form as
(2- 3)
A,
/"'-----~
x
Fig. 2-21
There is a distinct advantage to writing vectors in this manner.
Separating the magnitude and direction of each component vector will
simplify the operations of vector algebra, particularly in three dimensions.
z
A, k
I
A
i/
/ /
~----- Y
j
x
Fig. 2-22
Fig. 2-23
2.5
CARTESIAN VECTORS
z
Magnitude of a Cartesian Vector. If A is expressed as a Cartesian
vector, then its magnitude can be determined. As shown
in Fig. 2- 24, from the blue right triangle, A = VA 12 + Ai, and from
the gray right triangle, A = V Ai + A~. Combining these equations to
eliminate A ' yields
A
,k~I ,------..,.
1
I
A =
VA; + A~ + A~ I
(2-4)
x
/
Fig. 2-24
z
Hence, the magnilude of A is equal to the positive square root of the
sum of the squares of the magniludes of its components.
A,k
I
A
Coordinate Direction Angles. We will define the direction of A
by the coordinate direction angles a (alpha), f3 {beta), and y (gamma),
measured between the tail of A and the positive x, y, z axes, Fig. 2- 25.
Note that regardless of where A is directed, each of these angles will be
between 0° and 180°.
To determine a , {3, and y , consider the projection of A onto the x , y, z
axes, Fig. 2-26. Referring to the three shaded right triangles shown in the
figure, we have
Fig. 2-25
z
Ax
Ay
At
cos{3 = cosy = A
A
A
cos a = -
(2- 5)
These numbers are known as the direction cosines of A. Once they
have been obtained, the coordinate direction angles a , {3, y can then be
determined from the inverse cosines.
x
/
Fig. 2-26
41
42
CHAPTER
2
FORCE VECTORS
An easy way of obtaining these direction cosines is to form a unit
vector uA in the direction of A, Fig. 2- 25. To do this, divide A by its
magnitude A , so that
z
A
Ax
u = - = -j
I
A, k
A
A
A
A
Ay
At
+j + -k
A
A
(2-6)
By comparison with Eqs. 2- 5, it is seen that the i, j , k components of uA
represent the direction cosines of A , i.e.,
UA = COS
ai
+ COS {3j +
COS
yk
(2- 7)
Since the magnitude of uA is one, then from this equation an important
relation among the direction cosines can be formulated, namely,
l cos2 a
+ cos2 /3 + cos2 y
= 1
I
(2-8)
Therefore, if only two of the coordinate angles are known, the third
angle can be found using this equation.
Fmally, if the magnitude and coordinate direction angles of A are
known, then A may be expressed in Cartesian vector form as
A= AuA
= A cos a i + A cos /3j + A cos y k
= Axi + Ayj + Azk
(2- 9)
Horizontal and Vertical Angles. Sometimes the direction of A
z
I
can be specified using a horizontal angle() and a vertical angle </> (phi),
such as shown in Fig. 2- 27. The components of A can then be determined
by applying trigonometry first to the light blue right triangle, which yields
and
A ' = A sin</>
Now applying trigonometry to the dark blue right triangle,
Ax = A'cos () = A sin </> cos ()
Fig. 2-27
Ay = A 'sin() = A sin</> sin()
2.6
43
ADDITION OF CARTESIAN VECTORS
Therefore A written in Cartesian vector form becomes
A
= A sin </> cos (J i + A sin </> sin 8 j + A cos </> k
This equation should not be memorized; rather, it is important to
understand how the components were determined using trigonometry.
2. 6
ADDITION OF CARTESIAN
VECTORS
(A,+ BJk
The addition (or subtraction) of two or more vectors is greatly simplified
if the vectors are expressed in terms of their Cartesian components.
For example, if A = A..i + A1 j + Azk and B = B_.i + Byj + Bzk , Fig. 2-28,
then the resultant vector, R, has components which are the scalar sums of
the i, j , k components of A and B, i.e.,
R
= A + 8 = (A., + B., )i
(A1
+ B1 )j
- - -Y
+ (Ay + By)j + (Az + Bz)k
lf this is generalized and applied to a system of several concurrent
forces, then the force resultant is the vector sum of all the forces in the
system and can be written as
.r
Fig. 2-28
(2-10)
Here 'i.F_., 'i.F,, and 'i.F: represent the algebraic sums of the respective
x,y, z or i,j , k components of each force in the system.
IMPORTANT POINTS
• A Cartesian vector A has i, j, k components along the x, y,
axes. If A is known, its magnitude is A = YA} +A/ + A/.
z
• The direction of a Cartesian vector can be defined by the three
coordinate direction angles a, {3 , y , measured from the positive
x, y, z axes to the tail of the vector. To find these angEes,
formulate a unit vector in the direction of A , i.e., uA =A/ A,
and dete rmine the inverse cosines of its components. Only two
of these angles are independent of one another; the third angle
is found from cos2 a + cos2 f3 + cos 2 y = I.
The direction of a Cartesian vector can also be specified using
a horizontal angle (J and vertical angle</>.
Cartesian vector analysis provides a
convenient method for finding both the
resultant force and its components in three
dimensions.
44
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.7
Determine the magnitude and the coordinate direction angles of the
resultant force acting on the ring in Fig. 2- 29a.
z
F2 = {50i - lOOj + lOOk) lb F1 =160j + 80k}lb
x
x
(b)
(a)
Fig. 2-29
SOLUTION
Since each force is represented in Cartesian vector form, the resultant
force, shown in Fig. 2- 29b, is
FR= lF
=
F1 + F2
{60j + 80k } lb + {50i - lOOj + lOOk }lb
{50i - 40j + 180k }lb
=
=
The magnitude of FR is
+ (-40 lb) 2 + (180 lb)2
FR = Y(50 lb)2
=
=
191.0 lb
191 lb
Ans.
The coordinate direction angles a , {3, y are determined from the
components of the unit vector acting in the direction of FR·
FR
50 .
uF. = FR = 191.0
=
1
-
40 .
180 k
191.0J + 191.0
0.2617i - 0.2094j + 0.9422 k
so that
cos a = 0.2617
a = 74.8°
Ans.
cos {3
=
-0.2094
{3 = 102°
Ans.
cos y
=
0.9422
y = 19.6°
Ans.
These angles are shown in Fig. 2- 29b.
NOTE: Here {3
> 90° since the j component of
uF. is negative. This
seems reasonable considering how F 1 and F2 add according to the
parallelogram Jaw.
2.6
ADDITION OF CARTESIAN VECTORS
45
~ EXAMPLE 2.~
Express the force F shown in Fig. 2-30t1 as a Cartesian vector.
SOLUTION
The angles of 60° and 45° defining the direction of F are not coordinate
direction angles. Two successive applications of the parallelogram law
are needed to resolve F into its x, y, z components. FltSt F = F ' + F:,
then F ' = Fx + FP Fig. 2-30b. By trigonometry, the magnitudes of the
components are
=
F' =
Fx =
FY =
F:
F: 100 lb
= 86.6 lb
100 cos 60° lb = 50 lb
F' cos 45° = 50 cos 45° lb = 35.4 lb
F' sin 45° = 50 sin 45° lb = 35.4 lb
I 00 sin 60° lb
x
(a)
z
I
Realizing that F,, is in the - j direction, we have
F
= {35.4i -
35.4j + 86.6k } lb
Ans.
F: 100 lb
To show that the magnitude of this vector is indeed 100 lb, apply
Eq.2-4,
F = VF; + F; + F~
= v'C35.4)2 + (35.4)2 + (86.6)2 =
y
100 tb
F'
If needed, the coordinate direction angles of F can be determined from
the components of the unit vector acting in the direction of F.
x
(b)
u
F
F
Fx
j
F
35.4
=-j
-
=- =100
= 0.354i -
F,.
F.
+ -j + ~k
F
F
35.4
86.6
+ --k
100
100
--j
0.354j + 0.866k
F - lOOlb
so that
= cos- 1(0.354) = 69.3°
{3 = cos- (- 0.354) = 111°
'Y = cos- 1(0.866) = 30.0°
a
y
1
These results are shown in Fig. 2-30c.
x
(c)
Fig. 2-30
46
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.9
Two forces act on the hook shown in Fig. 2- 31a. Specify the magnitude of
F2 and its coordinate direction angles so that the resultant force FR acts
along the positive y axis and has a magnitude of 800 N.
z
SOLUTION
Fi
Y
To solve this problem, the resultant force FR and its two components,
F 1 and F 2, will each be expressed in Cartesian vector form. Then, as
shown in Fig. 2- 31b, it is necessary that FR = F1 + F2 .
Applying Eq. 2- 9,
+ Fi cos f3d + Fi cos y,k
Fi = F, cos a,i
x
(a)
F2
z
=
300 cos 45° i + 300 cos 60° j + 300 cos 120° k
=
{212.li + 150j - 150k}N
=
F2xi
+ F2yj + F2lk
Since FR has a magnitude of 800 N and acts in the +j direction,
FR
(800 N)( +j ) = { 800j } N
=
FR = F, + F2
800j = 212. li + 150j - 150k + F2xi + F2yj + F2zk
800j = (212.1 + F2x)i + (150 + F2y)j + (-150 + F2z)k
x
(b)
Fig. 2-31
To satisfy this equation the i, j , k components of FR must be equal to
the corresponding i,j, k components of (F1 + F 2). Hence,
+ F2x
150 + F2y
0 = 212.1
800 =
0 = -150 + F2t
F2r = -212.1 N
F2y
=
650 N
F2t
=
150 N
The magnitude of F 2 is thus
F2 = Y~(-2
-12.1-N
-)~
+_(_
650
_N
_)~
( 1_5_
0_
N~
)2
2_
2 _+_
700N
Ans.
We can use Eq. 2- 9 to determine a 2, 132, y 2•
-212.1
cos a2 = 700
Ans.
=
650
132 = 21.8°
/Jl
700'
150
COS"' = 'Y2 = 77.6°
12
700'·
These results are shown in Fig. 2- 31b.
cos
a _ = -·
Ans.
Ans.
2.6
A DDITION OF C ARTESIAN VECTORS
47
PRELIMINARY PROBLEMS
P2-3. Sketch the following forces on the x, y, z coordinate
axes. Show a , {3, y.
a) F = (50i + 60j - lOk} kN
b) F = {-40i - 80j
P2-S. Show how to resolve each force into its x, y, z
components. Set up the calculation used to find the
magnitude of each component.
z
+ 60k} kN
P2-4. In each case, establish F as a Cartesian vector, and
find the magnitude of F and the direction cosine of {3.
F= 600N
z
x
(a)
z
F=SOON
x
(a)
x
(b)
z
F= 800N
x
/
F
(b)
x
(c)
Prob. P2-4
Prob. P2-S
48
C HAPT ER
2
FORC E VECTORS
FUNDAMENTAL PROBLEMS
F2-13. Determine the coordinate direction angles of the
force.
F2-16. Express the force as a Cartesian vector.
z
Prob. F2-16
x
Express the force as a Cartesian vector.
FZ-17.
F= 75 lb
z
Prob. F2-13
F2-14. Express the force as a Cartesian vector.
z
F=500N
y
Prob.FZ-17
x
)'
FZ-18. Determine the resultant force acting on the hook.
z
Prob.FZ-14
F2-15. Express the force as a Cartesian vector.
z
)'
x
y
F= SOON
Prob. F2-15
Prob. F2-18
2.6
ADDITION OF CARTESIAN VECTORS
49
PROBLEMS
2-38. The force F has a magnitude of 80 lb. Determine the
magnitudes of the x, y, z components of F.
*2-40. Determine the magnitude and coordinate direction
angles of the force F acting on the support. The component
of Fin the x-y plane is 7 kN.
z
F,I
~----------~
F= 801b
- - - - - -y
7 kN
x
Prob. 2-40
Prob. 2-38
2-39. The bolt is subjected to the force F, which has
components acting along the x, y, z axes as shown. If the
magnitude of Fis 80 N, and a = 60° and 1' = 45°, determine
the magnitudes of its components.
2-41. Determine the magnitude and coordinate direction
angles of the resultant force and sketch this vector on the
coordinate system.
2-42. Specify the coordinate direction angles of F 1 and F2
and express each force as a Cartesian vector.
z
,r-- - - - . . , ; F,
F1 = 80 Jb
F2 = 130 lb
x
Prob. 2-39
I
Probs. 2-41142
50
CHAPTER
2
FORCE VECTORS
2-43. Express each force in Cartesian vector form and
then determine the resultant force. Find the magnitude and
coordinate direction angles of the resultant force.
2-47. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
z
*2-44. Determine the coordinate direction angles of F 1•
Fi= 125 N
Prob. 2-47
*2-48. Determine the magnitude and coordinate direction
angles of the resultant force, and sketch this vector on the
coordinate system.
z
Probs. 2-43/44
2-45. Determine the magnitude and coordinate direction
angles of F3 so that the resultant of the three forces acts
along the positive y axis and has a magnitude of 600 lb.
)'
2-46. Determine the magnitude and coordinate direction
angles of F3 so that the resultant of the three forces is zero.
F1 = 450N
Prob. 2-48
z
2-49. Determine the magnitude and coordinate direction
angles a t> 131> y 1 of F 1 so that the resultant of the three
forces acting on the bracket is FR= { - 350k } lb.
z
x
Fi= 200 lb
Fi= 300 lb
Probs. 2-45/46
x
Prob. 2-49
2 .6
2-50. If the resultant force FR has a magnitude of 150 lb
and the coordinate direction angles shown, determine the
magnitude of F2 and its coordinate direction angles.
51
A DDITION OF CARTESIAN V ECTORS
2-53. The spur gear is subjected to the two forces. Express
each force as a Cartesian vector.
2-54. The spur gear is subjected to the two forces.
Determine the resultant of the two forces and express the
result as a Cartesian vector.
y
Prob. 2-50
x
F1 = 50lb
Probs. 2-53/54
2-51. Express each force as a Cartesian vector.
*2-52. Determine the magnitude and coordinate direction
angles of the resultant force. and sketch this vector on the
coordinate system.
2-55. Determine the magnitude and coordinate direction
angles of the resultant force. and sketch this vector on the
coordinate system.
z
I
F2 =!SON
y
x
x
Probs. 2-51/52
Prob. 2-55
52
CHAPTER
2
FORCE VECTORS
2.7
POSITION VECTORS
In this section we will introduce the concept of a position vector. Later it
will be shown that this vector is of importance in formulating a Cartesian
force vector directed between two points in space.
B
(
4m
0 r2m-7
-,..=,1---~~-Y
4 .(
2 ril
'......L..
6m
lm
x
A
1
Fig. 2-32
x, y, z Coordinates. Throughout the book we will use the convention
followed in many technical books, which requires the positive z axis to be
directed upward (the zenith direction) so that it measures the height of
an object or the altitude of a point. Thex,y axes then lie in the horizontal
plane, Fig. 2- 32. Points in space are located relative to the origin of
coordinates, 0 , by successive measurements along the x, y, z axes. For
example, the coordinates of point A are obtained by starting at 0 and
measuring xA = +4 m along the x axis, YA = +2 m along they axis, and
finally ZA = -6 m along the z axis, so that A (4 m, 2 m, -6 m). In a similar
manner, measurements along the x, y, z axes from 0 to B give the
coordinates of B, that is, B(6 m , -1m, 4 m).
Position Vector. A position vector r is defined as a fixed vector
which locates a point in space relative to another point. For example, if r
extends from the origin of coordinates, 0, to point P(x, y, z), Fig. 2- 33a,
then r can be expressed in Cartesian vector form as
r =xi + yj + zk
Note how the head-to-tail vector addition of the three components yields
vector r, Fig. 2- 33b. Starting at the origin 0 , one "travels" x in the +i
direction, then y in the +j direction, and finally z in the +k direction to
arrive at point P(x, y, z).
z
z
P(x, y, z)
P(x, y, z)
r
r
yj
0
--Y
.\' i
xi
x
/
x
/
(a)
zk
0
yj
(b)
Fig. 2-33
--y
2. 7
POSITION V ECTORS
ln the more general case, the position vector may be directed
from point A to point B in space. From Fig. 2- 34a, by the headto-tail vector addition, using the triangle rule, we require
A(x,.. y,.. z,.)o
Solving for r and expressing rA and
r8
in Cartesian vector form yields
.t
(a)
or
{2-11)
Thus, the i, j , k components of r may be formed by taking the coordinates
of the lllil of the vector A(x11 , y11 , z11 ) and subtracting them from the
corresponding coordinates of the head B(x8 , y8 , z8 ) . We can also form
these components directly , Fig. 2-34b, by starting at A and moving
through a distance of (x 8 - xA) along the positive x axis (+i), then
(y8 - YA) along the positive y axis (+j ), and finally (z 8 - ZA) along the
positive z axis (+k) to get to B.
x
(b)
Fig. 2-34
U an x. y. z coordinate system is established,
B
then the coordinates of two points A and 8
on the cable can be determined. From this
the position vector r acting along the cable
can be formulated. Its magnitude represents
the distance from A to 8, and its unit vector,
u = r / r, gives the direction defined by a, {3. y.
53
54
I
2
CHAPTER
EXAMPLE
FORCE VECTORS
2.10
I
An elastic rubber band is attached to points A and B as shown in
Fig. 2- 35a. Determine its length and its direction measured from
A towardsB.
z
B~
2m ~ 3 m
;/i ~r
1
2m
y
Jm y
x
A~m
SOLUTION
We first establish a pos1t1on vector from A to B, Fig. 2- 35b. In
accordance with Eq. 2- 11, the coordinates of the tail A(l m, 0, -3 m)
are subtracted from the coordinates of the head B(-2 m, 2 m, 3 m),
which yields
r
(a)
=
[-2 m - 1 m] i
=
z
B
;1{6k{m
y
{
+
[2 m - O]j
+
[3 m - (-3 m)] k
-3i + 2j + 61k} m
These components of r can also be determined directly by realizing
that they represent the direction and distance one must travel along
each axis in order to move from A to B, i.e., along the x axis { -3i } m,
along they axis { 2j } m, and finally along the z axis { 6k } m.
The length of the rubber ba nd is therefore
r
r = V(-3 m)2
x
I
I 1- 3 iJm
A
(2 m) 2
+
(6 m)2 = 7 m
Ans.
{2j) m
Formulating a unit vector in the direction of r, we have
(b)
r
u=-=
r
B
•
-
3
7
-j
+
2
-j
7
+
6
7
- k
The components of this unit vector give the coordinate direction
angles
z'
r=7m
a =
'Y = 31.CY'
/3 =
a= 115°
+
cos- 1 (-~) =
73.4°
{3 = cos- 1(
A
x
(c)
Fi.g. 2-35
y
~)
115°
Ans.
73.4°
Ans.
= cos-{~) = 31.0°
Ans.
=
NOTE: These angles are measured from the posilive axes of a localized
coordinate system placed at the tail of r , as shown in Fig. 2- 35c.
2.8
2.8
F ORCE V ECTOR DIRECTED A LONG A LINE
FORCE VECTOR DIRECTED ALONG
A LINE
Quite often in three-dimensional statics problems, the direction of a force
is specified by rwo points through which its line of action passes. Such a
situation is shown in Fig. 2-36, where the force F is directed along the
cord AB. We can formuJate F as a Cartesian vector by realizing that it has
the same direction and sense as the position vector r directed from point
A to point B on the cord. This common direction is specified by the
unit vector u = r / r, and so once u is determined, then
x
Fig. 2.-36
Although we have represented F symbolically in Fig. 2- 36, note that it has
units offo rce, unlike r, which has units of length.
The force F acting along the rope can
be represented as a Cartesian vector by
establishing x, y. z axes and first forming a
•-
position vector r along the length of the
rope. Then the corresponding unit vector
u = r / r that defines the direction of both the
rope and the force can be determined. Finally.
tbe magnitude of tbe force is combin ed with
its direction. so tbat F = Fu.
IMPORTANT POINTS
• A position vector locates one point in space relative to
another point.
• The easiest way to formuJate the components of a position
vector is to determine the distance and direction that one must
travel in the x, y, z directions-going from the tail to the head
of the vector.
• A force F acting in the direction of a position vector r can be
represented in Cartesian form if the unit vector u of the position
vector is determined and it is muJtiplied by the magnitude of
the force, i.e., F = Fu = F(r / r).
55
56
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.11
The man shown in Fig. 2- 37a pulls on the cord with a force of 70 lb.
Represent this force acting on the support A as a Cartesian vector and
determine its direction.
SOLUTION
Force F is shown in Fig. 2- 37b. The direction of this vector, u, is
determined from the position vector r , which extends from A to B.
Rather than using the coordinates of the end points of the cord, r can
be determined directly by noting in Fig. 2- 37a that one must travel
from A {-24k} ft, then {-8j } ft , and finally {12i} ft to get to B. Thus,
r = { l 2i - 8j - 24k } ft
The magnitude of r , which represents the length of cord AB, is
r =
x
(a)
z'
V <12 ft)2 + <-8 ft)2 + <-24 ft)2 = 28 ft
Forming the unit vector that defines the direction and sense of both
r and F, we have
r
12.
8 .
24
u=-=-1 - -J - -k
r
28
28
28
Since F has a magnitude of 70 lb and a direction specified by u, then
F = Fu = 70 I'./.!3_i
'\28
=
{
- i_j 28
24
k)
28
30i - 20j - 60k } lb
Ans.
The coordinate direction angles are measured between the tail of r
(or F) and the positive axes of a localized coordinate system with
origin placed at A, Fig. 2- 37b. From the components of the unit vector:
r
·n
(b)
Fig. 2-37
cos-1(~~) =
64.6°
Ans.
{3 = cos- 1( : ) = 107°
2
Ans.
y = cos- 1( -24) = 149°
Ans.
a =
28
NOTE: These results make sense when compared with the angles
identified in Fig. 2- 37b.
2.8
-
EXAMPLE
FORCE VECTOR DIRECTED ALONG A LINE
2.12
-
The roof is supported by two cables as shown in the photo. If the cables
exert forces FA 8 = 100 N and FAc = 120 N on the wall hook at A as
shown in Fig. 2- 3&1, determine the resultant force acting at A. Express
the result as a Cartesian vector.
SOLUTION
The resultant force FR is shown graphically in Fig. 2- 38b. We can
express this force as a Cartesian vector by first formulating FAB and
FAC as Cartesian vectors and then adding their components. The
directions of FAB and FAC are specified by forming unit vectors u AB
and u Ac along the cables. These unit vectors are obtained from the
associated position vectors r AB and r Ac· With reference to Fig. 2- 3&t,
to go from A to B, we must travel { -4k } m, and then { 4i } m. Thus,
z
TAB = { 4i - 4k } m
rAB = V(4 m)2
+ (-4 m)2
=
5.66 m
( 4 .
4
)
TAB)
FAB = FAB ( rAB = (100 N) 5.661 - 5.66 k
FA8 = {70.7i-70.7k }N
To go from A to C, we must travel { -4k } m, then { 2j} m, and finally
{ 4i } m. Thus,
rAc
=
{
4i + 2j - 4k}
rAc = V(4 m)2
FAc =
=
m
(a)
z
+ (2 m) 2 + (-4 m) 2
FAc(~:~) =
{
x
(120 N)
=
6m
(~i + ~j - ~k)
80i + 40j - 80k } N
The resultant force is therefore
FR= FAB
+
FAc = {70.7i - 70.7k } N
- { 151i
+ {80i + 40j - 80k}
+ 40j - 151k} N
N
Ans.
x
(b)
Fig. 2-38
57
58
C HAPT ER
2
FORC E VECTORS
PRELIMINARY PROBLEMS
P2-6. In each case, establish a position vector from point
A to point B.
P2-7. In each case, express F as a Cartesian vector.
z
z
,,,--3 m - -7-,,,/
-:;;r-::,f"--- - --.-----,- Y
Sm
2m
~
A
_I
F= 15 kN
x
B
x
(a)
(a)
z
A
2m~
T
3m
lm
)'
+
J
y
4m
2m
lm
x
_i_
F= 600N
x
(b)
(b)
z
z
F= 300N
T
x
x
(c)
(c)
Prob. P2-6
Prob. P2-7
2.8
FORCE VECTOR DIRECTED A LONG A LINE
59
FUNDAMENTAL PROBLEMS
F2-19. Express rAs as a Cartesian vector, then determine its
magnitude and coordinate direction angles.
z
F2-22. Express the force as a Cartesian vector.
z
B
F= 900N
3m
.L fl:-- , , '
x
A
2m
Prob.F2-22
x
Prob.F2-19
F2-20. Determine the length of the rod and the position
vector directed from A to B. What is the angle 6?
F2-23.
at A.
Determine the magnitude of the resultant force
AZ~
6m
I
Fe= 420N
I
x
y
)'
x
Prob.F2-23
Prob.F2-20
F2-2L Express the force as a Cartesian vector.
F2-24. Determine the resultant force at A , expressed as a
Cartesian vector.
z
z
2m
2 ft
Fe= 490 lb
·~~t,: _ _:.:~_JCA'
I
F8 = 600lb
""&
r
/""!!> >
)'
x
x
4ft~ft
y
Prob.F2-21
Prob.F2-24
60
2
CHAPTER
FORCE VECTORS
PROBLEMS
*2-56. Determine the length of the connecting rod AB by
first formulating a position vector from A to B and then
determining its magnitude.
2-58. Express each force as a Cartesian vector, and then
determine the magnitude and coordinate direction angles
of the resultant force.
)'
z
c
/
B
F 1 = 801b
300mm
F2 = 50 lb
--.
, ~-h~~~~-P~~~ x
3(f
x
Pro b. 2-56
Prob. 2-58
2-57. Express force Fas a Cartesian vector; then determine
its coordinate direction angles.
z
A
/
/
2-59. If F = {350i - 250j - 450k } N and cable AB is
9 m long, determine the x, y, z coordinates of point A.
/
/
/
/
~~~~~~
~,..L~~~~..,,..:::::....~--'--"'----~~~~-y
,,.,
5 ft
x
x
Pro b. 2-57
Prob. 2-59
2.8
FORCE VECTOR D IRECTED ALONG A LINE
61
*2-60. The 8-m-long cable is anchored to the ground at A.
If x = 4 m and y = 2 m, determine the coordinate z to the
highest point of attachment along the column.
2-63. If Fe = 560 N and Fe = 700 N, determine the
magnitude and coordinate direction angles of the resultant
force acting on the flag pole.
2-6L The 8-m-long cable is anchored to the ground at A.
If z = 5 m, determine the location +x, +y of the support at A.
Choose a value such that x = y.
*2-64. If Fe = 700 N, and Fe = 560 N, determine the
magnitude and coordinate direction angles of the resultant
force acting on the flag pole.
z
z
)'
Probs. 2-63/64
x
2-65. The plate is suspended using the three cables which
exert the forces shown. Express each force as a Cartesian
vector.
Probs. 2-60/61
z
2-62. Express each of the forces in Cartesian vector form
and then determine the magnitude and coordinate direction
angles of the resultant force.
A
z
~AB= 250N
FAc = 400N
71
I
FcA = 5001b
3m
___./
___,__._..------;k:---,--~-,--
/
)'
FoA = 400 lb
2m
/c
B
,..Lt m
x
x
Prob. 2-62
Prob. 2-65
62
CHAPTER
2
FORCE VECTORS
2-66. Represent each cable force as a Cartesian vector.
"2-67. Determine the magnitude and coordinate direction
angles of the resultant force of the two forces acting at point A.
2-69. The load at A creates a force of 60 lb in wire AB.
Express this force as a Cartesian vector.
-~-Y
y
x
x
10 ft
x
Probs. 2-66167
Prob. "2-69
*2-68. The force F has a magnitude of 80 lb and acts at the
midpoint C of the rod. Express this force as a Cartesian
vector.
2-70. Determine the magnitude and coordinate direction
angles of the resultant force acting at point A on the post.
6 ft
3 ft
A
<.-<-- ----,<' - - - ' /-
,,L 2ft
x
x
Prob."2-68
Prob. 2-70
2. 9 Dor
PRODUCT
2.9 DOT PRODUCT
Occasionally in statics one has to find the angle between two lines or the
components of a force parallel and perpendicular to a line. In two
dimensions, these problems can readily be solved by trigonometry since
the geometry is easy to visualize. In three dimensions, however, this is
often difficult, and consequently vector methods should be employed for
the solution. The dot product, which defines a particular method for
"multiplying" two vectors, will be used to solve the above-mentioned
problems.
The dot product of vectors A and B, written A· B, and read " A dot B,"
is defined as the product of the magnitudes of A and B and the cosine of
the angle 6 between their tails, Fig. 2- 39. Expressed in equation form,
I A·B =
ABcose
I
(2- 12)
where 0° :$ 6 :$ 180°. The dot product is often referred to as the scalar
product of vectors since the result is a scalar and not a vector.
The following three Jaws of operation apply.
1. Commutative Jaw: A · B = B · A
2. Multiplication by a scalar: a(A · B) = (aA ) • B = A · (aB )
3. Distributive Jaw: A • (B + D) = (A · B) + (A · D)
Cartesian Vector Formulation. If we apply Eq. 2- 12, we can find
the dot product for any two Cartesian unit vectors. For ex.ample,
i • i = (1)(1) cos 0° = 1 and i • j = (1)(1) cos 90° = 0. If we want to find
the dot product of two general vectors A and B that are expressed in
Cartesian vector form, then we have
A · B = (A.,.i + A1 j + Azle) • (B) + Byj + Bzk )
= Afix (i · i) + A)3y (i · j ) + Afiz (i · k )
+ A>Bx (j · i) + (AyBy (j · j ) + A>Bz (j · k )
+ AzBx (k • i) + AzBy (k · j ) + AzBz (k • k)
Carrying out the dot-product operations, the final result becomes
(2- 13)
Thus, to determine the dot product of two Cartesian vectors, multiply their
corresponding x, y, z components and sum these products algebrt1ically.
The result will be either a positive or negative scalar, or it could be zero.
Fig. 2-39
63
64
CHAPTER
2
FORCE VECTORS
Applications. The dot product has two important applications.
• The angle formed b etween two vectors or intersecting lines. The
angle 6 between the tails of vectors A and B in Fig. 2- 39 can be
determined from Eq. 2- 12 and written as
Fig. 2-39 (Repeated)
6 =
(A·B)
cos- 1
AB
0° ::;
6 ::;
180°
Here A· B is found from Eq. 2- 13. As a special case, if A· B
then 6 = cos- 1 0 = 90° so that A will be perpendicular to B.
=
0,
• The components of a vector parallel and perpendicular to a line.
The component of vector A parallel to or collinear with the line aa
in Fig. 2-40 is defined by A 0 = A cos 6. This component is sometimes
referred to as the projection of A onto the line, since a right angle is
formed in the construction. If the direction of the line is specified by
the unit vector ua, and since u 0 = 1, we can determine the magnitude
of A 0 directly from the dot product (Eq. 2- 12); i.e.,
The angle () between the rope and the beam
can be determined by formulating unit
vectors along the beam and rope and then
using the dot product, u b • u, = ( 1)(1) cos 8.
Aa = A·ua = Acos6
Hence, the scalar projecti.on of A along a line is determined from the
dot product of A and the unit vector 0 0 which defines the direction of
the line. Notice that if this result is positive, then Aa has a directional
sense which is the same as u0 ; whereas if Aa is a negative scalar, then
Aa has the opposite sense of direction to ua.
The component A 0 represented as a vector is therefore
The perpendicular component of A can also be obtained, Fig. 2-40.
Since A = A 0 + A.L, then A.L = A - Aa .There are two possible ways
of obtaining A.i. One way would be to determine 6 from the dot product,
6 = cos- 1(A · uA/A); then A.i = A sin 6. Alternatively, if Aa is known,
then by the Pythagorean theorem we can also write A.i = V A2 - A~.
The projection of the cable force F along the
beam can be determined by first finding the
unit vector u b that defines this direction. Then
apply the dot product, Fb = F · ub.
A.=Acoso u.
Fig. 2-40
2. 9 Dor
PRODUCT
IMPORTANT POINTS
• The dot product is used to determine the angle between two
vectors or the projection of a vector in a specified direction.
• If vectors A and B are expressed in Cartesian vector form , the
dot product is determined by multiplying the respective x, y , z
scalar components and algebraically adding the results, i.e.,
A · B = AJ3.r + A>BY + AzBz·
• From the definition of the dot product, the angle formed
between the tails of vectors A and B is 8 = cos- 1 (A · B /AB).
• The magnitude of the projection of vector A along a line aa
whose direction is specified by 0 0 is determined from the dot
product, A0 = A · 0 0 .
I
EXAMPLE
2.13 1
Determine the magnitudes of the projections of the force F in Fig. 2-41
onto the u and v axes.
v
= lOON
50
Projections of F
(a)
Components of F
(b)
SOLUTION
The graphical representation of the projections
is shown in Fig. 2-41a. From this figure, the magnitudes of the
projections of F onto the u and v axes can be obtained by trigonometry:
Projections of Force.
(F,,)proj = (100 N)cos 45° = 70.7 N
Ans.
(Fv)proj = (100 N)cos 15° = 96.6 N
Ans.
These projections are not equal to the magnitudes of the
components of force F along the u and v axes found from the
parallelogram Jaw, Fig. 2-41b. They would only be equal if the u and
v axes were perpendicular to one another.
NOTE:
Fig. 2-41
65
66
I
CHAPTER
EXAMPLE
2
FORCE VECTORS
2.14
I
The frame shown in Fig. 2-42a is subjected to a horizontal force
F = {300j } N. Determine the magnitudes of the components of this force
parallel and perpendicular to member AB.
z
z
x
x
(a)
(b)
Fig. 2-42
SOLUTION
The magnitude of the projected component of F along AB is equal to
the dot product of F and the unit vector uB, which defines the direction
of AB, Fig. 2-42b. Since
rs
Us = -
rs
=
+ 6j + 3 k
Y( 2)2 + (6)2 + (3)2
2i
0.286i
+ 0.857j + 0.429 k
(300j ) · (0.286i
+ 0.857j + 0.429k)
=
then
FAS
=
F COS ()
=
(0)(0.286)
= 257.1 N
=
F ·Us
=
+ (300)(0.857) + (0)(0.429)
Ans.
Since the result is a positive scalar, FAB has the same sense of direction
as us, Fig. 2-42b.
Expressing FAB in Cartesian vector form, we have
FAS = FAsUs
= {73.5i
=
+
(257.1 N)(0.286i + 0.857j + 0.429k)
220j + llOk }N
Ans.
The perpendicular component, Fig. 2-43b, is therefore
F .i = F - FAs = 300j - (73.5i + 220j + 110k)
=
{
-73.5i
+ 79.6j - l lOk } N
Its magnitude can be determined either from this vector or by using
the Pythagorean theorem, Fig. 2-42b:
F.L =
F 2 - F1'8 = Y(300 N)2 - (257.1 N)2
= 155 N
Ans.
v
2. 9 Dor
-
EXAMPLE
PRODUCT
2.15
-
The pipe in Fig. 2-43a is subjected to the force of F = 80 lb. Determine
the angle 8 between F and the pipe segment BA, and the projection of
F along this segment.
c
x
(a)
SOLUTION
Angle 6.
First we will establish position vectors from B to A and B
to C; Fig. 2-43b. Then we will determine the angle 8 between the tails
of these two vectors.
r8 A = { -2i-2j + l k } ft , r8 A = 3 ft
rsc = { -3j + l k } ft, r8 c = v'iO ft
Thus,
x
cos B = r8A·rsc = (-2)(0)
rsArsc
+ (-2)(-3) + (1)(1)
3ViQ
=
0.
7379
8 = 42.5°
(b)
Ans.
The projection of F along BA is shown in
Fig. 2-43c. We must first formulate the unit vector along BA and force
F as Cartesian vectors.
Projection of F.
U 8A
r 8A
= -
F =
rsA
=
(-2i - 2j
3
+ l k)
2.
3
= - -1 -
801{~;~) = 8~- 3~lk) =
2.
1
-J + -k
3
3
-75.89j + 25.30k
x
Thus,
F8 A = F·u8 A = (-75.89j +
25.30k) · (-~i - ~j + ~k)
=
o(-~) + (-75.89\- ~) + (25.30)(~)
=
59.0 lb
NOTE: Since 8 has been calculated, then also, F8 A =
80 lb cos 42.5° = 59.0 lb.
Ans.
F cos 8 =
(c)
Fig. 2-43
67
68
CHAPTER
2
FORCE VECTORS
PRELIMINARY PROBLEMS
P2-8. In each case, set up the dot product to find the
angle e. Do not calculate the result.
P2-9. In each case, set up the dot product to find the
magnitude of the projection of the force F along a-a axes.
Do not calculate the result.
3 01
z
a
,,____
x
2m B
x
/2m
~..........,,
--F',~
__,.
lm
F=300N
_L
(a)
(a)
z
a
F= SOON
x
x
(b)
(b)
Prob. P2-8
Prob. P2-9
2. 9 Dor
69
PRODUCT
FUNDAMENTAL PROBLEMS
FZ-25. Determine the angle 8 between the force and the
line AO.
Find the magnitude of the projected component of
the force along the pipe AO.
FZ-29.
z
z
F=
1- 6 i + 9 j + 3 k) kN
x
)'
Prob.F2-25
Prob.FZ-29
FZ-26. Determine the angle 8 between the force and the
line AB.
Determine the components of the force acting
parallel and perpendicular to the axis of the pole.
F2-30.
z
z
-........n
4m
)'
x
Prob.F2-30
Prob.FZ-26
FZ-27. Determine the angle 8 between the force and the
line OA.
F2-3L Determine the magnitudes of the components of the
force F = 56 N acting along and perpendicular to line AO.
z
FZ-28. Determine the projected component of the force
along the line OA.
x
Probs. FZ-27/28
Prob.F2-31
70
CHAPTER
2
FORCE VECTORS
PROBLEMS
2-7L Given the three vectors A, B, and D, show that
A · (B
+
D)
= (A · B) + (A· D).
*2-72. Determine the magnitudes of the components of
F = 600 N acting along and perpendicular to segment DE
of the pipe assembly.
2-75. Determine the angle e between the two cables.
*2-76. Determine the magnitude of the projection of the
force F1 along cable AC.
B
~)'
x
y
F= 600N
"-'<"<:: - 3 m _ __,,
x
Probs. 2-75176
Probs. 2-7tn2
2-77. Determine the angle
wire AB.
2-73. Determine the angle e between BA and BC.
e between
2-74. Determine the magnitude of the projected
component of the 3 kN force acting along axis BC of the pipe.
z
:r
y
x
D
F= 3kN
Probs. 2-73n4
Prob. 2-77
the pole and the
2. 9 Dor
2- 78. Determine the magnitude of the projection of the
force along the u axis.
2-81.
71
PRODUCT
Determine the angle 8 between the two cables.
2-82. Determine the projected component of the force
acting in the direction of cable AC. Express the result as a
Cartesian vector.
z
F= 600N
c
v
r
2m y
x
8ft
10 ft
II
A
x
Prob. 2-78
Probs. 2-81182
Determine the angles 8 and <f> between the flag pole
and the cables AB and AC.
2-83.
2-79. Determine the magnitude of the projected component
of the 100-lb force acting along the axis BC of the pipe.
*2-80. Determine the angle 8 between pipe segments BA
and BC.
z
/
l.501
z
A
3~
x
~~'"
4:v ~
ft
• •• •• -- ··· ·
y""'
c--::F~
. 100 lb
Probs. 2-79/80
)'
x
y
Prob.2-83
72
CHAPTER
2
FORCE V ECTORS
*2-84. Determine the magnitudes of the components of F
acting along and perpendicular to segment BC of the pipe
assembly.
*2-88. D etermine the magnitudes of the components of
the force acting parallel and perpendicular to diagonal AB
of the crate.
2-85. Determine the magnitude of the projected
component of F along line AC. Express this component as a
Cartesian vector.
2~.
D etermine the angle () between the pipe segments
BA and BC.
Pro b. 2-88
2-89. Determine the magnitudes of the projected
components of the force acting along the x and y axes.
y
2-90. Determine the magnitude of the
component of the force acting along line OA.
projected
Probs. 2-84185/86
x
2-87. If the force F= 100 N lies in the plane DBEC. which
is parallel to the x- z plane. and makes an angle of 10" with
the extended line DB as shown. determine the angle that F
makes with the diagonal AB of the crate.
Probs. 2-89/90
2-91. Two cables exert forces on the pipe. Determine the
magnitude of the projected component of F1 along the line
of action of F 2•
*2-92.
Determine the angle() between the two forces.
z
x
F 1 = 30 lb
Prob. 2-87
Pro bs. 2-91192
73
CHAPTER REVlEW
CHAPTER REVIEW
A scalar is a positive or negative number;
e.g., mass and temperature.
A vector has a magnitude and direction.
where the arrowhead represents the
sense of the vector.
Multiplication or division of a vector by a
scalar will change only the magnitude of
the vector. If the scalar is negative, the
sense of the vector will change so that it
acts in the opposite direction.
If vectors are co llinear, the resultant is
simply the algebraic or scalar addition.
R -
A+ B
A
B
Parallelogram Law
Two forces add according to the
parallelogram law. The components form
the sides of the parallelogram and the
resu/10111 is the diagonal.
a
\
To find the components of a force along
any two axes, extend lines Crom the head
of the force. parallel to the axes, to form
the components.
Two force components can be added
tip-to-tail using the triangle rule, and
then the law of cosines and the law of
sines can be used to calculate unknown
values.
Resultant
c----
-
-
Components
FR= V F T + F~ - 2F1F2cos8R
F2
FR
-Fi- =--=--
-
- -1
74
CHAPTER
2
FORCE VECTORS
Re dangular Components: 1\vo Dimensions
y
Vectors F, and F,. are rectangular components
ofF.
F,
The resultant force is determined from the
algebraic sum of its components.
y
F1.r
(FR\ = 'i.F,
(FR)y = 'i.F>,
FR = V(FR);
8 = tan- I
y
F2x
~,,, ,,
-
·1
,,
+ (FR);
F11
t
FJ)'
,.,,...-'
,'
..
F.lT
7
.. 1 Ft.r
' '~
(FR)y
Cartesian Vectors
The unit vector u has a length of 1, no units, and
it points in the direction of the vector F.
u
F
F
= -
A force can be resolved into its Cartesian
components along the x, y, z axes so that
F = F) + F,J + F)r..
z
F, k
F
The magnitude of F is determined from the
positive square root of the sum of the squares of
its components.
The coordinate direction angles a , /3. 1' are
determined by formulating a unit vector in the
direction of F. The x, y, z components of
u represent cos a , cos /3, cos 1'·
I
F =VF}+ F} + F}
F
F,
F,.
F,
f.k
u
= F = y i + Fj
u
= cosa i + cosf3 j + cosy k
+
-
)1
CHAPTER REVIEW
The coordinate direction angles are
related. so that only two of the three
angles arc independent of one another.
75
coi1 a + cos2 f3 + cos2 1' = 1
To find the resultant of a concurrent force
system. express each force as a Cartesian
vector and add the i.j , k components of all
the forces in the system.
Position and Force Vectors
A position vector locates one point in space
relative to another. The easiest way to
formulate the components of a position
vector is to determine the distance and
direction that one must travel along the
x, y, and z directions-going from the tai l to
the head of the vector.
+
(yB - YA)j
+
(l/i - ZA)k
11::,_---J:=:i==~=::.....-y
(yB - YA)j
F =Fo =F(~)
If the line of action of a force passes
through points A and 8, then the force
acts in the same direction u as the
position vector r extending from A to 8.
Knowing F and u. the force can then be
expressed as a Cartesian vector.
x
Dot Product
The dot product between two vectors A
and B yields a scalar. If A and B arc
expressed in Cartesian vector form, then
the dot product is the sum of the products
of their x,y, and z components.
The dot product can be used to determine
the angle between A and B.
The dot product is also used to
determine the projected component of a
vector A onto an axis aa defined by its
unit vector u•.
A· B =
ABcos8
= A,Bx + A/Jy + A:!J,
8 = cos-
A.
A· B)
( AB
1
= A cos 8 u,, = (A • u )u
0
a
0
A 0 = A cos 0
Da
--.
- a
00
76
CHAPT ER
2
FORC E VECTORS
REVIEW PROBLEMS
R2-1. Determine the magnitude of the resultant force FR
and its direction, measured clockwise from the positive
u axis.
R2-3. Determine the magnitude of the resultant force
acting on the gusset plare.
/
Fi= 200 lb
x
Prob. R2-1
/
/
F4 = 300 lb
s
4
F3 = 300 lb
Prob. R2-3
R2-2. Resolve the force into components along the u and
v axes and determine the magnitudes of these components.
*R:Z-4. The cable exerts a force of 250 lb on the crane
boom as shown. Express this force as a Cartesian vector.
z
v
F= 250N
/
~"
F= 250lb
Prob. R2-2
Prob. R2-4
REVIEW PROBLEMS
R2-5. The cable attached to the tractor at B exerts a force
of 350 lb on the framework. Express this force as a Cartesian
vector.
77
R2-7. Determine the angle 8 between the edges of the
sheet-metal bracket.
1-400
z
mm _ _ _ _J
l
x _ __
y
x
Prob. R2-7
Prob. R2-5
*R2-8. Determine the projection of the force F along
the pole.
R2~.
Express F 1 and F2 as Cartesian vectors.
F = l2i + 4j
Prob. R2-6
Prob. R2-8
+ tot } kN
CHAPTER
3
(©Rolf Adlercreutz/Alamy)
The force applied to this wrench will produce rotation or a tendency for rotation.
This effect is called a moment, and in this chapter we will study how to determine
the moment of a system of forces and calcu late their resultants.
FORCE SYSTEM
RESULTANTS
CHAPTER OBJECTIVES
•
To discuss the concept of the moment of a force and show how
to calculate it in two and three dimensions.
•
To provide a method for finding the moment of a force about a
z
specified axis.
•
To define the moment of a couple.
•
To show how to find the resultant effect of a nonconcurrent
force system.
•
(a)
z
To indicate how to reduce a simple distributed loading to a
resultant force acting at a specified location.
3.1
MOMENT OF A FORCE-SCALAR
FORMULATION
When a force is applied to a body it will produce a tendency for the body to
rotate about a point that is not on the line of action of the force. This tendency
to rotate is sometimes called a torque, but most often it is called the moment
of a force or simply the moment. For example, consider applying a force to
the handle of the wrench used to unscrew the bolt in Fig. 3-la. The magnitude
of the moment is directly proportional to the magnitude of F and the
perpendicular distance or moment arm d. The larger the force or the longer
the moment arm, the greater the moment or turning effect. If the force F is
applied at an angle(} ¥ 90°, Fig. 3-lb, then it will be more difficult to turn
the bolt since the moment arm d' = d sin(} will be smaller than d. If F is
applied along the handle, Fig. 3-lc, its moment arm will be zero since the line
of action of F will intersect point 0 (the z axis). As a result, the moment of
F about 0 is also zero and no turning can occur.
(b)
z
(c)
Fig. 3-1
79
80
CHAPTER
3
FORCE SYSTEM RESULTANTS
In general, if we consider the force F and point 0 to lie in the shaded
plane shown in Fig. 3-2a, the moment Mo about point 0, or about an
axis passing through 0 and perpendicular to the plane, is a vector quantity
since it has a specified magnitude and direction.
Magnitude. The magnitude of M 0 is
(3- 1)
F
(a)
Sense of rotation
where d is the moment arm or perpendicular distance from the axis at
point 0 to the line of action of the force. Units of moment magnitude
consist of force times distance, e.g., N · m or lb· ft.
Direction. The direction of Mo is defined by its moment axis, which is
(b)
Fig. 3-2
perpendicular to the plane that contains the force F and its moment arm d.
The right-hand rule is used to establish the sense of direction of M 0 , where
the natural curl of the fingers of the right hand, as they are drawn towards the
palm, represents the rotation, or the tendency for rotation caused by the
moment. Doing this, the thumb of the right hand will give the directional sense
of M 0 , Fig. 3-2a. Here the moment vector is represented three-dimensionally
by a curl around an arrow. In two dimensions this vector is represented only
by the curl, as in Fig. 3-2b. Since this produces counterclockwise rotation, the
moment vector is actually directed out of the page.
Resultant Moment. For two-dimensional problems, where all the forces
lie within thex- y plane, Fig. 3-3, the resultant moment (MR)o about point 0
(the z axis) can be determined by finding the algebraic sum of the moments
caused by all the forces in the system. As a convention, we will generally
consider positive moments as counterclockwise since they are directed along
the positive z axis (out of the page). C/.ockwise moments will be negative.
The directional sense of each moment can be represented by a plus or minus
sign. Using this sign convention, with a symbolic curl to define the positive
direction, the resultant moment in Fig. 3-3 is therefore
Fig. 3-3
If the numerical result of this sum is a positive scalar, (MR)owill be a
counterclockwise moment (out of the page); and if the result is negative,
(MR)owill be a clockwise moment (into the page).
3.1
-
EXAMPLE
3.1
MOMENT OF A FORCE-
81
SCALAR FORMULATION
-
For each case illustrated in Fig. ~, determine the moment of the force
about point 0.
SOLUTION (SCALAR ANALYSIS)
The line of action of each force is extended as a dashed line in order to
establish the moment arm d. Also illustrated is the tendency of rotation
of the member as caused by the force, and the orbit of the force about 0
is shown as a colored curl. Thus,
Fig. 3-4a
M 0 = (100N)(2m) = 200N · m)
Ans.
Fig. 3-4b
M 0 = (50 N)(0.75 m) = 37.5 N · m)
Ans.
Fig. 3-4c
Mo = (40 Jb)(4 ft+ 2 cos 30° ft) = 229 Jb · ft)
Ans.
Fig. 3-4d
Mo = (60 Jb)(l sin 45° ft) = 42.4 lb · ft )
Ans.
Fig. 3-4e
Mo = (7 kN)(4 m - 1 m) = 21.0 kN · m)
Ans.
lOON
!
~)
0
2m
(a)
2ttA
:~401b
q(o\.)
0 f:=t==:;;;;;;;;;;;;;;;;;o~~ 1
I
1--4 f t - - - 1 - - -:
0.75 m
1~'---- SON
2 cos 30" ft
(b)
(c)
1-201-I
'
lm
i - - - - -3 ft
0
~--'-- 7kN
----11
I)_----_'ili·.'·"r,."
4m
601b
(d)
0
Fig. 3-4
(e)
82
I
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.2
Determine the resultant moment of the four forces acting on the rod
shown in Fig. 3- 5 about point 0.
SOLUTION
y
Assuming that positive moments act m the + k direction, 1.e.,
counterclockwise, we have
SON
2m !~
- - -
60N
_;...;3m- ~,;;,...- 20N
C+ (MR)o
=
2Fd;
(MR)o
=
- 50 N(2 m) + 60 N(O) + 20 N(3 sin 30° m)
- 40 N(4 m + 3 cos 30° m)
"-/
(MR) 0
=
- 334 N · m
=
334N · m
;>
Ans.
40N
For this calculation, note how the moment-arm distances for the 20-N
and 40-N forces are established from the extended (dashed) lines of
action of each of the forces.
Fig. 3-5
F
The force F tends to rotate the beam clockwise about its
support at A with a moment MA = FdA. The actual rotation
would occur on ly if the support at B were removed.
The ability to remove the nail will require
the moment of F,, about point 0 to
be larger than the moment of the force
F,, about 0 that is needed to pull the
nail out.
3.2
3. 2
CROSS PRODUCT
The moment of a force will be formulated using Cartesian vectors in the
next section. Before doing this, however, it is first necessary to expand our
knowledge of vector algebra and introduce the cross-product method of
vector multiplication.
The cross p roduct of two vectors A and B yields the vector C, which
is wrinen as
C=AX B
(3-2)
and is read "C equals A cross B."
Magnitude. The magnitude of C is defined as the product of the
magnitudes of A and B and the sine of the angle 8 between their tails,
where 0° < fJ s 1800. Thus,
C =AB sin 8
Direction . Vector Chas a direction that is perpendicular to the plane
containing A and B such that the directional sense of C is specified by
the right-hand rule; i.e., curling the fingers of the right hand from
vector A (cross) to vector B, the thumb points in the direction of C, as
shown in Fig. 3~.
Knowing both the magnitude and direction of C, we can therefore write
C
=A x
B
=
(A B sin 8)u c
The terms of Eq. 3-3 are illustrated graphically in Fig. 3-6.
C= AXB
B
Fig. 3-6
(3-3)
CROSS PRODUCT
83
84
CHAPTER
3
FORCE SYSTEM RESULTANTS
The following three Jaws of operation apply.
• The commutative Jaw is not valid; i.e., A x B
C= A XB
~
B
x
A. Rather,
A x B =-B x A
This is shown in Fig. 3-7 by using the right-hand rule. The cross
product B x A yields a vector that has the same magnitude but acts
in the opposite sense of direction to C; i.e., B x A = -C.
8
A
•
If the cross product is multiplied by a scalar a, it obeys the associa-
tive Jaw;
a(A x B) = (aA ) x B
=
A x (aB) = (A x B)a
-C = B X A
Fig. 3-7
This property is easily shown since the magnitude of the resultant
vector (I aIAB sin 8) and its sense of direction are the same in each
case.
• The vector cross product also obeys the distributive Jaw of addition,
z
A
x (B + D)
=
(A
x B) + (A x D)
k = i Xj
It is important to note that proper order of these cross products
must be maintained since they are not commutative.
Cartesian Vector Formulation. Equation 3- 3 may be used to
find the cross product of any pair of Cartesian unit vectors. For example,
to find i x j , the magnitude of the resultant vector is
(i)(j)(sin 90°) = (1)(1)(1) = 1, and its direction is determined using the
right-hand rule, Fig. 3-8. Here the resultant vector points in the + k
direction so that i x j = (l)k. In a similar manner,
x/
Fig. 3-8
i X j =k
j X k=i
k X i=j
Fig. 3-9
i x k= -j
i X i=O
j X i= - k j X j =0
k x j - -i k X k=O
These results should not be memorized; rather, it should be clearly
understood how each is obtained by using the right-hand rule and the
definition of the cross product. A simple scheme shown in Fig. 3- 9 can
sometimes be helpful for obtaining the same results when the need
arises. If the circle is constructed as shown, then "crossing" two unit
vectors in a counterclockwise fashion around the circle yields the positive
third unit vector; e.g., k x i = j. "Crossing" clockwise, a negative unit
vector is obtained; e.g., i x k = - j.
3.2
Let us now consider the cross product of vectors A and B which are
expressed in Cartesian vector form. We have
A X B
=
(Axi + A,,j + Azk) x (B..i + B,,j + Bzk)
- A..B..(i x i) + A..B,,(i x j ) + A..Bz(i x k)
+ A,, B..(j x i) + A,,B,,(j x j ) + A,,Bz(j x k)
+ A:B..(k x i) + A, B,,(k x j ) + AzBz(k x k)
Carrying out the cross-product operations and combining terms yields
This equation may also be written in a more compact determinant
form as
j
k
(3-5)
Thus, to find the cross product of A and B, it is necessary to expand a
determinant whose first row of elements consists of the unit vectors i,j ,
and k and whose second and third rows represent the x, y, z components
of the two vectors A and B, respectively.*
•A determinant having three rows and three columns can be expanded using three minors,
each of which is multiplied by one of the three terms in the first row. There are four
elements in each minor. for example.
By definition. this determinant notation represents the terms (A 11A 22 - A 1i}l21), which is
simply the product of the two elements intersected by the arrow slanting downwar d to the
right (A 11 A22) mi1111s the product of the two elements intersected by the arrow slanting
downward to the left (A 1;021 ). For a 3 X 3 determinant. such as Eq. 3-5. the three minors
can be generated in accordance with the following scheme:
fo• olom"I lo
m-
l(A,B, -,.....
A_,B
_>_.) _ ___,
?
Fo"lomoo< j
Fo r element k:
.
'
~
B
',
B,
~
1
Remember the
negative sign
-j(A,B, - A ,B, )
~! = k(A,8
1 -
AyB,)
Adding these results and noting that the j element must include 1he mi1111s sign yields the
expanded form of A x B given by Eq. ~-
CROSS PRODUCT
85
86
CHAPT ER
3
FORC E SYST EM RESU LTANTS
3.3 MOMENT OF A FORCE-VECTOR
FORMULATION
The moment of a force F about point 0 , Fig. 3- lOa, can be expressed using
the vector cross product,
Moment axis
Mo = r x F
(3-6)
Here r is a position vector directed from 0 to any point on the line of
action of F. We will now show that indeed the moment M 0 , when
determined by this cross product, has the proper magnitude and direction.
Magnitude. The magnitude of the cross product is defined from
Eq. 3- 3 as M0 = rF sin 8, where the angle 8 is measured between the
F
(a)
tails of r and F . To establish this angle, r must be treated as a sliding
vector so that 8 can be constrUJcted properly, Fig. 3- lOb. Since the moment
arm d = r sin 8, then
Moment axis
t
~ dQ
Mo
0
F
(b)
Fig. 3-10
_. t)
M 0 = r 1 X F = r2 X F = r3 X F
I
Mo = rFsin 8
=
F(rsin 8)
=
Fd
which agrees with Eq. 3- 1.
Direction. The direction an d sense of M o in Eq. 3- 6 are determined
by the right-hand rule as it applies to the cross product. Thus, slidi ng
r to the dashed position and curling the right-hand fingers from r
towards F, "r cross F," the thumb is directed upward or perpendicular
to the plane containing r and F and this is in the same direction as
M0 , the moment of the force about point 0 , Fig. 3- lOb. Remember
that the cross product does not obey the commutative Jaw, and so the
order of r x F must be mai ntained to produce the correct sense of
direction for M 0 .
Principle of Transmissibility. The cross product operation is often
used in three dimensions since the perpendicular distance or moment
arm from point 0 to the line of action of the force is not needed. In other
words, we can use any position vector r measured from point 0 to any
point on the line of action of the force F, Fig. 3- 11. Thus,
0
F
M0
Line of action
Fig. 3-11
=
r1 x F
=
r2 x F
=
r3 x F
Since F can be applied at any point along its line of action and still create
this same moment about point 0, then F can be considered a sliding
vector. This property is called the principle of transm issibility of a force.
3.3
Cartesian Vector Formulation. If we establish x , y, z coordinate
axes, then the position vector r and force F can be expressed as Cartesian
vectors, Fig. 3-12a. Then applying Eq. 3- 5 we have
M o= r x F
=
i
j
k
rx
ry
r-
F,,
F, E
Moment
axis\
F
Mo\ ~
y
(3-7)
x
/
(a)
where
r,., r,,, r:
87
MOMENT OF A FORCE-VECTOR FORMULATION
represent the x, y, z components of the position
vector drawn from point 0 to any point on the
line of action of the force
Fx, F,,, Fz represent the x,y, z components of the force vector
If the determinant is expanded, then like Eq. 3-4 we have
- -- Y
(b}
The physical meaning of these three moment components becomes
evident by studying Fig. 3-12b. For example, the i component of M o
can be d etermined from the moments of Fx, F,,, and Fi about the x axis.
The component Fx does not create a moment or tendency to cause
turning about the x axis since this force is parallel to the x axis. The line
of action of F,. passes through point B , and so the magnitude of the
moment of F,, about po int A on the x axis is r~F,· B y the right-hand rule
this component acts in the negative i direction. Likewise, Fz passes
through point C and so it contributes a moment component of r,.Fz i
about the x axis. Thus, (Mo)x = (r,.F,. - r4 F,) as shown in Eq. 3-8. As
an exercise, try to establish the j and k components of M 0 in this
manner and show that indeed the expanded form of the determinant,
Eq. 3-8, represents the moment of F about point 0. Once M o is
determined, realize that it will always be perpendicular to the shaded
plane containing vectors r and F, Fig. 3- 12a.
Fig. 3-12
F,
Resultant Moment of a System of Forces. If a body is acted
upon by a system of forces, Fig. 3-13, the resultant moment of the forces
about point 0 can be determined by vector addition of the moment of
each force. This resultant can be written symbolically as
/
.r
{3-9)
Fig. 3-13
88
I
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.3
z
Determine the moment produced by the force Fin Fig. 3-14a about
point 0. Express the result as a Cartesian vector.
SOLUTION
As shown in Fig. 3-l4b, either rA or rs can be used to determine the
moment about point 0. These position vectors are
r A = {12k} m
rs = {4i
+ 12j} m
Force F expressed as a Cartesian vector is
F
= Fu As =
2kN
x
=
{4i + 12j - 12k} m
]
[ V(4m) 2 + (12m)2 + (-12m) 2
{0.4588i + 1.376ji - 1.376k} kN
Thus
(a)
M o = rA X F =
=
i
0
j
k
0
0.4588
1.376
12
-1.376
(0(-1.376) - 12(1.376)]i - (0(-1.376) - 12(0.4588)]j
+ (0(1.376) - 0(0.4588)]k
=
{-16.5i + 5.5lj} kN · m
Ans.
or
i
M 0 =rs
x
xF
=
4
0.4588
j
12
1.376
k
0
-1.376
=
(12(-1.376) - 0(1.376)]i - (4(-1.376) - 0(0.4588)]j
+ (4(1.376) - 12(0.4588)]k
=
{-16.5i + 5.5lj} kN · m
Ans.
NOTE: As shown in Fig. 3- 14b, M o acts perpendicular to the plane
that contains F, r A, and r s. H ad this problem been worked using
(b)
Fig. 3-14
Mo = Fd, notice the difficulty that would arise in obtaining the
moment arm d.
3.3
I
EX AMPLE
MOMENT OF A FORCE- V ECTOR FORMULATION
3.4
Two forces act on the rod shown in Fig. 3-15a. Determine the resultant
moment they create about the flange at 0. Express the result as a
Cartesian vector.
F 1 = {- 60i
+ 40j + 20k} lb
0
x
B
F2 = {80i
+ 40j - 30k} lb
(b}
(a)
SOLUTION
Position vectors are directed from 0 to each force as shown m
Fig. 3-15b. These vectors are
TA =
{5j } ft
r8
{4i + 5j - 2k} ft
=
Moment
axis
The resultant moment about 0 is therefore
(MR)o
=
I(r
i
=
x
(c)
x F1 + r8 x F2
= rA
-
X F)
j
5
40
k
i
j
5
40
k
0
0 + 4
-2
- 60
20
80
-30
[5(20) - 0(40)Ji - [O]j + (0( 40) - (5)(-60)Jk
Fig. 3-15
+ [5(-30) - (-2)(40)J i - (4(- 30) - (- 2)(80)J j + (4(40) - 5(80)Jk
=
{30i - 40j + 60k} lb . ft
Ans.
NOTE: This result is shown in Fig. 3- 15c. The coordinate direction
angles were determined from the unit vector for (MR) 0 . Realize that
the two forces tend to cause the rod to rotate about the moment axis
in the manner shown by the curl indicated on the moment vector.
89
90
CHAPTER
3
FORCE SYSTEM RESULTANTS
3.4
r
0
PRINCIPLE OF MOMENTS
A concept often used in mechanics is the principle of moments, which is
sometimes referred to as Varignon's theorem since it was originally
developed by the French mathematician Pierre Varignon (1654-1722). It
states that the moment of a force about a point is equal to the sum of the
moments of the components of the force about the point. This theorem can
be proven easily using the vector cross product since the cross product obeys
the distributive law. For example, consider the moments of the force F and
twoofitscomponentsaboutpointO, Fig.3-16.Since F = F1 + F2 wehave
Fig. 3-16
M0 = r
I
x F
I
=
r x (F1 + F2)
moment of force
=
r x F1 + r x F 2
moment of components
For two-dimensional problems, Fig. 3- 17, we can use the principle of
moments by resolving the force into any two rectangular components
and then determine the moment using a scalar analysis. Thus,
M 0 = F.y-F.x
x
y
The following examples will show that this method is generally easier
than finding the same moment using M0 = Fd.
0
Fig. 3-17
IMPORTANT POINTS
• The moment of a force creates the tendency of a body to turn
about an axis passing through a specific point 0.
• Using the right-hand rule, the sense of rotation is indicated by the
curl of the fingers, and the thumb produces the sense of direction
of the moment.
• The magnitude of the moment is determined from M 0 = Fd,
where d is called the moment arm, which represents the
perpendicular or shortest distance from point 0 to the line of
action of the force.
• In three dimensions the vector cross product is used to determine
the moment, i.e., M o = r x F. Here r is directed from point 0 to
any point on the line of action of F.
• In two dimensions it is often easier to use the principle of
The moment of the force about point 0 is
M o = Fd. But it is easier to find this moment
using Mo = f.(O) + Fyr = Fyr.
moments and find the moment of the force's components about
point 0 , rather than using M 0 = Fd.
3.4
I
EXAMPLE
PRINCIPLE OF MOMENTS
3.5
Determine the moment of the force in Fig. 3- 18a about point 0.
y
-
d, = 3cos30°m - I
F, = (5 kN) cos 45°
---"""A"7~
I
d1 = 3 sin 30" m
I
0
(a)
(b)
SOLUTION I
The moment arm din Fig. 3- 18a can be found from trigonometry.
d = (3 m) sin 75° = 2.898 m
Thus,
M 0 = Fd = (5 kN)(2.898 m) = 14.5 kN · m)
Ans.
Since the force tends to rotate or orbit clockwise about point 0 , the
moment is directed into the page.
SOLUTION II
The x and y components of the force are indicated in Fig. 3- 18b.
Considering counterclockwise moments as positive, and applying the
principle of moments, we have
C+ M 0
=
-Frdy - f'ydx
=
-(5 cos 45° kN)(3 sin 30° m) - (5 sin 45° kN)(3 cos 30° m)
-14.5 kN · m = 14.5 kN · m )
Ans.
=
F, = (5 kN) cos 75°
y
SOLUTION Ill
The x and y axes can be oriented parallel and perpendicular to the
rod's axis as shown in Fig. 3-18c. Here~. produces no moment about
point 0 since its line of action passes through this point. Therefore,
C+ Mo =
-F,,dx
0
=
-(5 sin 75° kN)(3 m)
=
-14.5kN·m
=
14.5kN · m)
(c)
Ans.
Fig. 3-18
91
92
I
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.6
Force Facts at the end of the angle bracket in Fig. 3-19a. Determine
the moment of the force about point 0.
SOLUTION I (SCALAR ANALYSIS)
The force is resolved into its x and y components, Fig. 3-19b, then
(. + M 0
400 sin 30° N(0.2 m) - 400 cos 30° N(0.4 m)
=
=
-98.6 N · m
=
98.6 N · m )
or
M o = {-98.6k} N · m
Ans.
SOLUTION II (VECTOR ANALYSIS)
Using a Cartesian vector approach, the force and position vectors,
Fig. 3-19c, are
r = {0.4i - 0.2j } m
F
=
(400 sin 30°i - 400 cos 30°j) N
=
{200.0i - 346.4j} N
The moment is therefore
y
i
0.4
200.0
j
-0.2
- 346.4
k
=
r x F
0.2m
=
Oi - Oj + (0.4(-346.4) - (-0.2)(200.0)Jk
~i---~--"'V.-1
=
{-98.6k} N · m
M0
=
0
0
Ans.
0.4 m _ J \
~
(c)
Fig. 3-19
F
NOTE: The scalar analysis (Solution I) provides a more convenient
method for analysis than Solution II since the direction of the
moment and the moment arm for each component force are easy to
establish. For this reason, this method is generally recommended for
solving problems in two dimensions, whereas a Cartesian vector
analysis is generally recommended only for solving threedimensional problems.
3.4
PRINCIPLE OF MOMENTS
93
PRELIMINARY PROBLEMS
P3-L In each case, determine the moment of the force
about point 0.
l
, _
~
;fi-1
F = l- 3i
+ 2j + Sk )kN
SOON
(a)
(g)
lOON
I
x
(a)
11---3m----1
0 lr-== = = = = i - ,
2
1-lm- I
3m ----1
' l __
lm
_JJ_
1-tm-I
(b)
(h)
SOON
t~
o ------(',Jf
I
__.I. J 1:m
2 m--il
-
l - - 3m -
'k
z
0
lOON
I
P3-2. In each case, set up the determ inant to find the
moment of the force about point P.
z
SOON
2m - I
lm
(c)
_l_
1- lml -2m- l - -3m- - I
s
4
-I
x
2m
0
3
0
SOON
F = l2i - 4j - 3k ) kN
1
(b)
(i)
(d)
;1·~\1--_-_-_-_- -m==--=--=--=-~__,I
-S
z
O
F = {- 2i + 3j
+ 4k} kN
lOON
(e)
O
~===~=l::iO~O=N==J
l-2m ~1-- 3m -J
(f)
x
(c)
Prob. P3-1
Prob. P3-2
94
CHAPTER
3
FORCE SYSTEM RESULTANTS
FUNDAMENTAL PROBLEMS
F3-1. Determine the moment of the force about point 0.
F:>-4. Determine the moment of the force about point 0.
lOON
~
=1
2m
_I
0
\ 1---Sm - - -
1
Prob.F3-l
Prob. F:>-4
F3-2. Determine the moment of the force about point 0.
F3-S. Determine the moment of the force about point 0.
6001b
F=300N
0
~0.4m--I
Prob.F3-2
\
F3-3. Determine the moment of the force about point 0.
Prob.F3-S
F3-6. Determine the moment of the force about point 0.
4ft -I /~
SOON
0
6001b
Prob.F3-3
Prob.F3-6
3.4
F3-7. Determine the resultant moment produced by the
forces about point 0.
95
PRINCIPLE OF MOMENTS
£3-10. Determine the moment of force F about point 0.
Express the result as a Cartesian vector.
SOON
0
y
? 1b 13-.
l3-11. Determine the moment of force F about point 0.
Express the result as a Cartesian vector.
600N
Prob. FJ-7
FJ-8. Determin e th e resultant moment produced by the
forces about point 0.
F= 1201b
B
r2 fl
F 1 = 500N
.............y
x
I
Yr• • >-11
0.25m
1
0
Prob. FJ-8
• 3-1<. Determine the resultant moment produced by the
forces about point 0.
F2 = 200 lb
.3-U. If
F1 = POOi - 120j + 75kJ lb
and
F2 =
(-200i + 250j + IOOkJ lb, determine the resultant moment
produced by these forces about point 0. Express the result
as a Cartesian vector.
z
6f1--
I
30°
F 1 = 300 1b
Prob. F3-9
y
Prob. FJ-12
96
3
CHAPTER
FORCE SYSTEM RESULTANTS
PROBLEMS
3-L Prove the distributive Jaw for the vector cross
product, i.e., A x (B + D) = (A x B) + (A x D).
3- 7. Determine the moment of each of the three forces
about point A.
3-2. Prove the triple
A· (B x C) =(A x B)-C.
*3-8. Determine the moment of each of the three forces
about point B.
scalar
product
identity
3-3. Given the three nonzero vectors A, B, and C, show
that if A · (B x C) = 0, the three vectors must lie in the
same plane.
*3-4. Determine the moment about point A of each of the
three forces acting on the beam.
F 1 =250N 3Cl°
2 m- J - - -3 m- -+---1
3-5. Determine the moment about point B of each of the
three forces acting on the beam.
4m
F2 = 500 lb
F 1 = 375 lb
A
8ft-+-6ft+5ft
Probs. 3-718
F3 = 160 lb
Probs. 3-4/5
3-9. Determine the moment of each force about the bolt
at A. Take F8 = 40 lb, Fe = 50 lb.
3-6. The crowbar is subjected to a vertical force of P = 25 lb
at the grip, whereas it takes a force of F = 155 lb at the claw
to pull the nail out. Find the moment of each force about
point A and determine if P is sufficient to pull out the nail.
3-10. If F8 = 30 lb and Fe = 45 lb, determine the
resultant moment about the bolt at A.
p '
,;? -
14 in.
Prob.3-6
Probs. 3-9/10
3.4
3-11. The cable exerts a force of P = 6 kN at the end
of the 8-m-long crane boom. If 8 = 300. determine the
placement x of the hook at B so that this force creates a
maximum moment about point 0. Whal is this moment?
*3-U . The cable exerts a force of P = 6 kN at the end of
the 8-m-long crane boom. If x = 10 m, determine the angle 8
of the boom so that this force creates a maximum moment
about point 0. What is this moment?
97
PRINCIPLE OF MOMENTS
3-15. Two men exert forces of F = 80 lb and P = SO lb on
the ropes. Determine the moment of each force about A.
Which way will the pole rotate, clockwise or counterclockwise?
*3-16. If the man at B exerts a force of P = 30 lb on the
rope, determine the magnitude of the force F the man at C
must exert to prevent the pole from rotating. i.e., so the
resultant moment about A of both forces is zero.
A
I
6rt
_I
F
12 rt
'-- - -x·- - -- -
c
Probs. 3-11/12
A
Probs. 3-15116
3-13. The 20.N horizontal force acts on the handle of the
socket wrench. What is the moment of this force about point B.
Specify the coordinate direction angles a, {3, 'Y of the moment
axis.
3-14. 111e 20-N horizontal force acts on the handle of the
socket wrench. Determine the moment of this force about
point 0. Specify the coordinate direction angles a , {3, y of
the moment axis.
3-17. ' Ille torque wrench ABC is used to measure the
moment or torque applied to a bolt when the bolt is located
at A and a force is applied to the handle at C. The mechanic
reads the torque on the scale at B. If an e:>.1ension AO of
length dis used on the wrench, determine the required scale
reading if the desired torque on the bolt at 0 is to be M.
20N
(
B
/
200mm
F
/11
y
~
0
1-d
Probs. 3-13114
A
B
s~
I
Prob. 3-17
4
c
98
CHAPTER
3
F ORCE SYSTEM RESULTANTS
3-18. The tongs are used to grip the ends of the drilling pipe.
Determine the torque (moment) Mp that the applied force
F = 150 lb exerts on the pipe about point P as a function of 8.
Plot this moment Mp versus 8 for (J' s 8 < 90".
3-19. The tongs arc used to grip the ends of the drilling
pipe. If a torque (moment) of MP = 800 lb· ft is needed
at P to turn the pipe. determine the cable force F that must
be applied to the tongs. Set 8 = 30°.
F
3-22. Old clocks were constructed using a fusee B to drive
the gears and watch hands. The purpose of the fusee is to
increase the leverage developed by the mainspring A as it
uncoils and thereby loses some of its tension. The mainspring
can develop a torque (moment) T, = k8, where
k = O.OlS N · m/rad is the torsional stiffness and 8 is the
angle of twist of the spring in radians. If the torque T1
developed by the fusee is to remain constant as the
mainspring winds down. and x = 10 mm when 8 = 4 rad,
determine the required radius of the fusee when 8 = 3 rad.
y
12mm
~~
-..,..
IT,
1- - -- - 4 3 in. ------1
Probs. 3-18/19
*3-20. The handle of the hammer is subjected to the force
of F = 20 lb. Determine the moment of this force about the
point A.
3-21. In order to pull out the nail at B, the force F exerted
on the handle of the hammer must produce a clockwise
moment of 500 lb· in. about point A. Determine the
required magnitude of force F.
Prob. 3-22
3-23. The tower crane is used to hoist the 2-Mg load upward
at constant velocity. The LS-Mg jib BD. 0.S-Mg jib BC, and
6-Mg counterweight C have centers of mass at Gi. G2. and G3,
respectively. Determine the resultant moment produced by
the load and the weights of the tower crane jibs about point A
and about point B.
*3-24. The tower crane is used to hoist a 2-Mg load upward
at constant velocity. The LS-Mg jib BD and 0.S-Mg jib BC
have centers of mass at G 1 and G2. respectively. Determine
the required mass of the counterweight C so that the
resultant moment produced by the load and the weight of
the tower crane jibs about point A is zero. The center of mass
for the counterweight is located at G3.
A-9.sm-i
As
G2 I
I
O:::::mzszs:zm;~i=--
12.5 m~
23 Ill
A
Probs. 3-20/21
Probs. 3-23124
D
3.4
3-25. If the 1500-lb boom AB. the 200-lb cage BCD, and
the 175-lb man have centers of gravity located at points G 1,
G 2 , and G3 , respectively, determine the resultant moment
produced by each weight about point A.
3-26. If the 1500-lb boom AB, the 200-lb cage BCD, and
the 175-lb man have centers of gravity loca ted at points GI>
G 2, and G3 , respectively, determine the resultant moment
produced by all the weights about point A.
99
PRINCIPLE OF M OMENTS
3-29. The force F = {400i - lOOj - 700k) lb acts at the
end of Che beam. Determine the moment of this force about
point 0.
3-30. TI1e force F = {400i - lOOj - 700k} lb acts at the
end of t he beam. Determine the moment of this force about
point A.
5 ft 1.75 ft
Probs. 3-29/30
Probs. 3-25/26
3-27. Determine the moment of the force F about point 0.
Express the result as a Cartesian vector.
3-31. Determine the moment of the force F about point P.
Express the result as a Cartesian vector.
• 3-28. Determine the moment of the force F about point P.
Express the result as a Cartesian vector.
p
F = {-6i + 4 j
2L
+ 8k } kN
2Z'
4mlp
y
3m
I3 m
6m
J
lm
y
0
3 111
y
F = {2i
+ 4j - 6k} kN
Im
x
x
Probs. 3-27128
Prob. 3-31
3m
1 00
C H APT ER
3
FORC E SYST EM RESU LTANTS
*3-32. The curved rod has a radius of 5 ft. If a force of
60 lb acts at its end as shown, determine the moment of this
force about point C.
3-33. Determine the smallest force F that must be applied
along the rope in order to cause the curved rod to fail at the
support C. This requires a moment of M = 80 lb· ft to be
developed at C.
3-35. The pipe assembly is subjected to the 80-N force.
Determine the moment of this force about point A.
*3-36. The pipe assembly is subjected to the 80-N force.
Determine the moment of this force about point B.
z
A
z
x
60 Jb
6ft
U
,<--------aB
x
--~~-7ft ~~-?~,A
Probs. 3-35/36
Probs. 3-31133
3-34. A 20-N horizontal force is applied perpendicular to
the handle of the socket wrench. Determine the magnitude
and the coordinate direction angles of the moment created
by this force about point 0.
z
I
------
3-37. A force of F = {6i - 2j + l k} kN produces a
moment of M 0 = {4i + Sj - 14k} kN · m about the origin,
point 0. If the force acts at a point having an x coordinate of
x = 1 m, determine the y and z coordinates. Nore: The
figure shows F and M 0 in an arbitrary position.
3-38. The force F = {6i + 8j + lOk} N creates a moment
about point 0 of M 0 = {- 14i + 8j + 2k }N · m. If the
force passes through a point having an x coordinate of 1 m,
determine t he y and z coordinates of the point. Also,
realizing that M0 = Fd, determine the perpendicular
distanced from point 0 to the line of action of F. Nore: The
figure shows F and M 0 in an arbitrary position.
/
z
/
F
200mm
/
/
/
Mo
75mm
L~~-1~
/
'&
z
0 ,,,,...__,,__--=:::::::::;~
lm
y
y
x
Prob.3-34
y
0
Probs. 3-37/38
3.5
3.5
MOMENT OF A FORCE ABOUT A SPECIFIED AxlS
101
MOMENT OF A FORCE ABOUT
A SPECIFIED AXIS
Sometimes the moment produced by a force about a specified axis must
be determined. For example, suppose the lug nut at 0 on the car tire in
Fig. 3- 20a needs to be loosened. The force applied to the wrench will
create a tendency for the wrench and the nut to rotate about the moment
axis passing through 0; however, the nut can only rotate about they axis.
Therefore, to determine the turning effect, only they component of the
moment is needed, and the total moment produced is not important. To
determine this component, we can use either a scalar or vector analysis.
Scalar Analysis
To use a scalar analysis, the moment arm, or
perpendicular distance d0 from the axis to the line of action of the force,
must be determined. The moment is then
(3-10)
For example, for the lug nut in Fig. 3-20a, dy = d cos 8, and so the moment
of F about they axis is M,, = F d,, = F(d cos 8). According to the righthand rule, My is directed along the positive y axis as shown in the figure.
y
(a)
Fig. 3-20
1 02
CHAPTER
3
FORCE SYSTEM RESULTANTS
Vector Analysis. To find tbe moment of force F in Fig. 3-20b about the
x
Y
y axis using a vector analysis, we must first determine the moment of the force
about any point 0 on the y axis by applying Eq. 3- 7, M 0 = r x F.
The component My along they axis is the projection of M o onto they axis. It
can be found using the dot product discussed in Chapter 2, so that
My = j ·Mo = j · (r x F), where j is the unit vector for they axis.
We can generalize this approach by Jetting Ua be the unit vector that
specifies the direction of the a axis, shown in Fig. 3- 21. Then the moment
of F about a point 0 on the axis is M 0 = r x F, and the projection of this
moment onto the a axis is M 0 = ua · (r x F). This combination is referred
to as the scalar triple product. If the vectors are written in Cartesian
form, we have
(b)
Fig. 3-20 (cont.)
Ma
=
=
(ua} + Ua1 j +
i
Ua,k ] · rx
j
ry
k
rz
F.r
F,.
Fz
Ua,(ryFz - rzF,,) - u 0 1(rxFz - rzF.r) + ua.(rxF,• - ryFx)
This result can also be written in the form of a determinant, making it
easier to memorize.*
a
Ma
= 0
0 • (
r X F) =
Ua,
rx
(3- 11)
F.r
where
Uax ' Ua y 'Ua;
rx, ry, rz represent the x, y , z components of the position
vector extended from any point 0 on the a axis to
any point A on the line of action of the force
f ,, Fy, Fz represent the x,y, z components of the force vector.
r
F
Axis of projection
Fig. 3-21
represent the x, y, z components of the unit
vector defining the direction of the a axis
When Ma is evaluated from Eq. 3- 11, it will yield a positive or negative
scalar. The sign of this scalar indicates the sense of direction of Ma
along the a axis. If it is positive, then Ma will have the same sense as u°'
whereas if it is negative, then Ma will act opposite to u 0 . Once the a axis
is established, point your right-hand thumb in the direction of Ma, and
the curl of your fingers will indicate the sense of twist about the axis,
Fig. 3- 21.
*Take a moment to expand this determinant, to show that it will yield the above result.
3.5
MOMENT OF A FORCE ABOUT A SPECIFIED AxlS
Once Mu is determined, we can then express Ma as a Cartesian vector,
namely,
(3-12)
The examples which follow illustrate numerical applications of these
concepts.
IMPORTANT POINTS
• The mome nt of a force about a specified axis can be determined
provided the perpendicular distance da from the force line of
action to the axis can be determined. Then Ma = Fda.
• If vector analysis is used, then Ma = ua • (r X F), where 0 0
defines the direction of the axis and r is extended from any point
on the axis to any point on the line of action of the force.
• If M 0 is calculated as a negative scalar, then the sense of direction
of Mu is opposite to 0 0 .
• The moment M 11 expressed as a Cartesian vector is determined
from M 0 = M11u11 •
EXAMPLE
3.;-i
Determine the resultant moment of the three forces in Fig. 3-22 about
the x axis. they axis, and the z axis.
SOLUTION
A force that is parallel to a coordinate axis or bas a line of action that
passes through the axis does nor produce any moment or tendency for
turning about the axis. Defining the positive direction of the moment
of a force according to the right-hand rule, as shown in the figure,
we have
= (601b)(2ft)
+ (50lb)(2ft) + 0
=
220lb·ft
Ans.
My = 0 - (50 lb)(3 ft) - (40 lb)(2 ft)
=
-230 lb · ft
Ans.
Mx
M,
=0+0-
(401b)(2ft)
=
-80lb · ft
Ans.
The negative signs indicate that My and M z act in the -y and
directions, respective ly.
-z
Fig. 3-22
1 03
1 04
-
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.8
Determine the moment M AB produced by the force F in Fig. 3-23a,
which tends to rotate the rod about the AB axis.
SOLUTION
A vector analysis using MAB = u 8 · (r x F) will be considered for the
solution rather than trying to find the moment arm or perpendicular
distance from the line of action of F to the AB axis.
Unit vector u 8 defines the direction of the AB axis of the rod,
Fig. 3- 23b, where
F=300N
r8
Ua = -
ra
{0.4i + 0.2J m
0
}
=
V(0.4 m)2 + (0.2 m)2
=
0.8944i + 0.4472j
(a)
x
Vector r is directed from any point on the AB axis to any point on the
line of action of the force. For example, position vectors re and r0 are
suitable, Fig. 3-23b. (Although not shown, r 8 c or r80 can also be
used.) For simplicity, we choose r0 , where
z
r o = {0.6i} m
The force is
A - - - -Y
-
F
=
{-300k} N
Substituting these vectors into the determinant form of the triple
product and expanding, we have
F
MAB = u a · (r o X
F)
(b)
Fig. 3-23
=
=
0.8944
0.6
0
0.4472
0
0
0
0
-300
0.8944(0(-300) - 0(0)) - 0.4472(0.6(-300) - 0(0))
+ 0(0.6(0) - 0(0))
80.50N·m
This positive result indicates that the sense of
direction as u 8 , Fig. 3- 23b.
Expressing MAB as a Cartesian vector yields
=
M AB = MABU B =
=
MAB
is in the same
(80.50 N · m)(0.8944i + 0.4472j)
{72.0i + 36.0j} N · m
Ans.
NOTE: If the axis AB is defined using a unit vector directed from
B toward A , then in the above determinant - u 8 would have to
be used. This would lead to MAB = -80.50 N · m. Consequently,
M AB = MA 8 (-u 8 ) , and the same vector result would be obtained.
3.5
-
EXAMPLE
3.9
MOMENT OF A FORCE ABOUT A SPECIFIED
1 05
Axis
-
Determine the magnitude of the moment of force F about segment OA
of the pipe assembly in Fig. 3- 24a.
z
~
SOLUTION
0.5 m
D~
The moment of F about the OA axis is determined from
MoA = u 0 A • (r X F), where r is a position vector extending from
any point on the OA axis to any point on the line of action of F. /,~
As indicated in Fig. 3- 24b, either r 0 D, r 0 c, r AD• or r Ac can be used.
Here r oD will be considered since it will simplify the calculation.
The unit vector u 0 A , which specifies the direction of the OA axis, is O.Sm
{0.3i + 0.4j} m
r oA
U QA
= -
=
roA
Y(0.3 m) 2 + (0.4 m) 2
=
.
.
0.61 + 0.8J
x
and the position vector roD is
(a)
r oD = {0.5i
+ 0.5k} m
The force F expressed as a Cartesian vector is
F
= F(r cD)
rcD
=
=
(300N)
{0.4i - 0.4j + 0.2k} m
[
V (0.4 m) 2 +
]
(-0.4 m) 2 + (0.2 m) 2
y
x
{200i - 200j + lOOk } N
(b)
Therefore,
Fig. 3-24
MoA = u oA · ( r oD X F)
-
0.6
0.5
200
0.8
0
-200
0
0.5
100
=
0.6(0(100) - (0.5)(-200)) - 0.8(0.5(100) - (0.5)(200)] + 0
=
lOON·m
Ans.
1 06
C H APT ER
3
FORC E SYST EM RESU LTANTS
PRELIMINARY PROBLEMS
P3-3. In each case, determine the resultant moment of the
forces acting about the x, they, and the z axis.
P3-4. In each case, set up the determinant needed to find
the moment of the force about the a-a axes.
F = 16i
+ 2j + 3k ) kN
z
200N
ISON
4m
300N
x
a
(a)
(a)
3 n1
,..___------<-u
4 m __""/
f
/ --
/
2m
_l
x
F = {2i - 4j
+ 3k} kN
(b)
(b)
F = {2i - 4j
+ 3k ) kN
z
SON
I
lm-1
7
300N
3 n1
400N
lOON
(c)
Prob. P3-3
(c)
Prob. P3-4
3.5
1 07
M OMENT OF A FORCE ABOUT A SPECIFIED Axis
FUNDAMENTAL PROBLEMS
t"J-13. D etermine the magnitude of the moment of the
force F = ( 300i - 200j + 1SOk J N about the x axis.
."3-14. Determine the magnitude of the moment of the
force F = (300i - 200j + 150k ) N about the OA axis.
• 3-16. Determine the magnitude of the moment of the
force about the y axis.
F = (30i - 20j
+ 50kl N
Prob. Fl-16
t '3-17. Determine
the
moment
of
the
force
F = (50i - 40j + 20k ) lb about the AB axis. Express the
result as a Cartesian vector.
x
0.2111
F
Probs. Fl-13114
• 3-15. Determine the magnitude of the moment of the
200·N force about the x axis. Solve the problem using both a
scalar and a vector analysis.
.-r•
--
0.3m--
F=200N
>-.
1'3-1 •• Determine the moment of force F about the x. the
y, and the z axis. Solve the problem using both a scalar and
a vector analysis.
z
F= SOON
0.25111
y
Prob. 1''3-15
Prob. 13-18
1 08
CHAPTER
3
FORCE SYSTEM RESULTANTS
PROBLEMS
3-39. The lug nut on the wheel of the automobile is to be
removed using the wrench and applying the vertical force of
F = 30 Nat A. Determine if this force is adequate, provided
14 N · m of torque about the x axis is initially required to
turn the nut. If the 30-N force can be applied at A in any
other direction, will it be possible to turn the nut?
3-43. Determine the magnitude of the moment of the force
F about the x, they, and the z axis. Solve the problem (a) using
a Cartesian vector approach and (b) using a scalar approach.
*3-44. Determine the moment of force F about an axis
extending between A and C. Express the result as a
Cartesian vector.
*3-40. Solve Prob. 3-39 if the cheater pipe AB is slipped
over the handle of the wrench and the 30-N force can be
applied at any point and in any direction on the assembly.
z
z
F= 30N
y
)'
2 ft
in
x
F = l4i
Probs. 3-39/40
3-4L The A-frame is being hoisted into an upright
position by the vertical force of F = 80 lb. Determine the
moment of this force about the y' axis passing through
points A and B when the frame is in the position shown.
3-42. The A-frame is being hoisted into an upright
position by the vertical force of F = 80 lb. Determine the
moment of this force about the x axis when the frame is in
the position shown.
z
+ 12j - 3k) lb
Probs. 3-43/44
3-45. The board is used to hold the end of the cross lug wrench
in the position shown when the man applies a force of
F = 100 N. Determine the magnitude of the moment produced
by this force about the x axis. Force Flies in a vertical plane.
3-46. The 'board is used to hold the end of the cross lug
wrench in the position shown. If a torque of 30 N · m about
thex axis is required to tighten the nut, determine the required
magnitude of the force F needed to turn the wrench. Force F
lies in a vertical plane.
z
F
250mm
y
x'
)'
x
250mm
x
Probs. 3-41142
Probs. 3-45/46
3.5
3-47. The A-frame is being hoisted into an upright position
by the vertical force of F = 80 lb. Determine the moment
of this force about the y axis when the frame is in the
position shown.
1 09
MOMENT OF A FORCE ABOUT A SPECIFIED AxlS
3-51. A horizontal force of F = {-SOi} N is applied
perpendicular to the handle of the pipe wrench. Determine
the moment that this force exerts along the axis OA (z axis)
of the pipe assembly. Both the wrench and pipe assembly,
OABC, lie in the .r-z plane. Suggestion: Use a scalar analysis.
*3-52. Determine the magnitude of the horizontal force
F = - F i acting on Lbe handle of the pipe wrench so that
this force produces a component of the moment along the
OA axis (z axis) of the pipe assembly of M, = {4k} N · m.
Both the wrench and the pipe assembly, OABC, lie in
the .r-z plane. Suggestion: Use a scalar analysis.
F
x'---<l' ~
y
6ft~8
x
I
Prob.3-47
*3-48. Determine the magnitude of the moment of the force
F = {SOi - 20j - 80k} N about member AB of the tripod.
3-49. Determine the magnitude of the moment of the force
F = {SOi - 20j - 80k} N about member BC of the tripod.
3-50. Determine the magnitude of the moment of the force
F = {SOi - 20j - 80k) N about member CA of the tripod.
0
x
)'
Probs. 3-51/52
3-53. D etermine the moment of the force about the a- a
axis of the pipe if a = 6'1'. f3 = 6'1'. and y = 45°. Also.
determine the coordinate direction angles of Fin order to
produce the maximum moment about the a-a axis. What is
this moment?
z
F=30N
~y
a
~
IOOmm
a
y
Pro bs. 3-48/49/50
Pro b. 3-53
11 0
CHAPTER
3
FORCE SYSTEM RESULTANTS
3.6
A couple is defined as two parallel forces that have the same magnitude,
but opposite directions, and are separated by a perpendicular distanced,
Fig. 3-25. Since the resultant force is zero, the only effect of a couple is to
produce a rotation, or if no movement is possible, there is a tendency for
rotation.
The moment produced by a couple is called a couple moment. We can
determine its value by finding the sum of the moments of both couple
forces about any arbitrary point. For example, in Fig. 3- 26, position
vectors r A and r8 are directed from point 0 to points A and B lying on
the line of action of -F and F. The couple moment determined about 0
is therefore
F
-F
Fig. 3-25
A
F
0
Fig. 3-26
MOMENT OF A COUPLE
-F
However, r8
= rA
+ r or r
=
r8
-
r A, so that
M
=
r x F
(3- 13)
This result indicates that a couple moment is a free vector, i.e., it can
act at any point since M depends only upon the position vector r directed
between the forces and not the position vectors rA and r8 , directed from
point 0 to the forces.
Scalar Formulation. The moment of a couple, Fig. 3- 27, has a
magnitude of
(3-14)
where Fis the magnitude of one of the forces and d is the perpendicular
distance or moment arm between the forces. The direction and sense of
the couple moment are determined by the right-hand rule, where the
thumb indicates this direction when the fingers are curled with the sense
of rotation caused by the couple forces. In all cases, M will act
perpendicular to the plane contairLing these forces.
Vector Formulation. As noted above, the moment of a couple can
also be expressed by the vector cross product using Eq. 3- 13, i.e.,
(3-15)
Fig. 3-27
Application of this equation is easily remembered if one thinks of taking
the moments of both forces about a point lying on the line of action of
one of the forces. For example, if moments are taken about point A in
Fig. 3- 26, the moment of-Fis zero about this point, and the moment of
Fis defined from Eq. 3- 15. Therefore, in the formulation r is crossed with
the force F to which it is directed.
3.6
MOMENT OF A COUPLE
Fig. 3-28
Equivalent Couples. If two couples produce a moment with the same
magnitude and direction, then these two couples are equivalent. For example,
the two couples shown in Fig. 3-28 are equivalent because each couple
moment has a magnitude of M = 30 N(0.4 m) = 40 N(0.3 m) = 12 N · m,
and each is directed into the plane of the page. Notice that larger forces are
required in the second case to create the same turning effect because the
hands are placed closer together. Also, if the wheel was connected to the
shaft at a point other than at its center, then the wheel would still tum when
each couple is applied since this 12 N · m couple is a free vector.
Resultant Couple Moment. Since couple moments are free
vectors, their resultant can be determined by moving them to a single
point and using vector addition. For example, to find the resultant of
couple moments M1 and M2 acting on the pipe assembly in Fig. 3- 29a,
we can join their tails at point 0 and find the resultant couple moment,
MR = M 1 + M 2, as shown in Fig. 3-29b.
If more than two couple moments act on the body, we may generalize
this concept and write the vector resultant as
(a)
(3- 16)
These concepts are illustrated numerically in the examples that follow.
In general, problems projected in two dimensions should be solved using
a scalar analysis since the moment arms and force components are easy
to determine.
MR
(b)
Fig. 3-29
111
112
CHAPTER
3
FORCE S YSTEM RESULTANTS
IMPORTANT POINTS
• A couple moment is produced by two noncollinear forces that
are equal in magnitude but opposite in direction. Its effect is to
produce pure rotation, or tendency for rotation in a specified
direction.
• A couple moment is a free vector, and as a result it causes the
same rotational effect on a body regardless of where the couple
moment is applied to the body.
Steering wheels on vehicles have been
made smaller than on older vehicles
because power steering does not require
the dri ver to appl y a large couple
moment to the wheel.
• The moment of the two couple forces can be determined about
any point. For convenience, this point is often chosen on the
line of action of one of the forces in order to eliminate the
moment of this force about the point.
• In three dimensions thle couple moment is often determined
using the vector formulation, M = r x F, where r is directed
from any point on the line of action of one of the forces to any
point on the line of action of the other force F.
• A resultant couple moment is simply the vector sum of all the
couple moments of the system.
I
EXAMPLE
3.10
I
F1 = 2001b
F3 = 300 lb
Determine the resultant couple moment of the three couples acting on
the plate in Fig. 3-30.
-d1=4 ft-
F2 = 450 Jb A
SOLUTION
As shown the perpendicular distances between each pair of couple forces
are d 1 = 4 ft , d 2 = 3 ft, and d 3 = 5 ft. Considering counterclockwise
couple moments as positive, we have
F 1 = 200 lb
F3 = 300 lb
Fig. 3-30
=
-(200 lb)(4 ft) + (450 lb)(3 ft) - (300 lb)(5 ft)
=
-950 lb . ft
=
950 lb . ft
;>
Ans.
3.6
i
EXAMPLE
MOMENT OF A COUPLE
3.11
Determine the magnitude and direction of the couple moment acting on
the gear in Fig. 3-31a.
F= 600N
F = 600 N
(a)
600 sin 30" N
(b)
SOLUTION
The easiest solution requires resolving each force into its components
as shown in Fig. 3- 31b. The couple moment can be determined by
summing the moments of these force components about any point, for
example, the center 0 of the gear or point A. If we consider
counterclockwise moments as positive, we have
<: + M
=
2.M0 ; M
=
(600 cos 30° N)(0.2 m) - (600 sin 30° N)(0.2 m)
=
43.9N·m)
=
(600 cos 30° N)(0.2 m) - (600 sin 30° N)(0.2 m)
=
43.9N·m )
Ans.
or
<: + M
=
2.MA; M
Ans.
This positive result indicates that M has a counterclockwise rotational
sense, so it is directed outward, perpendicular to the page.
NOTE: The same result can also be obtained using M = Fd, where dis
the perpendicular distance between the lines of action of the couple
forces, Fig. 3- 31c. However, the computation for d is more involved.
Also, realize that the couple moment is a free vector and can act at any
point on the gear and produce the same turning effect about point 0.
F= 600N
(c)
Fig. 3-31
11 3
114
I
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.12
I
Determine the couple moment acting on the pipe shown in Fig. 3-32a.
Segment AB is directed 30° be[ow the x- y plane.
z
z
x
x
z
(b)
(a)
SOLUTION I (VECTOR ANALYSIS)
The moment of the two couple forces can be found about any point. If
point 0 is considered, Fig. 3- 32b, we have
x
A
----y
M
=
=
=
(c)
z
x (-25k) + r 8 x (25k)
(8j) X (-25k) + (6 cos 30°i + 8j - 6 sin 30°k)
-200i - 129.9j + 200i
{-130j} lb· in.
= rA
X
(25k)
Ans.
It is easier to take moments of the couple forces about a point lying on
the line of action of one of the forces, e.g., point A , Fig. 3- 32c. In this
case the moment of the force at A is zero, so that
M
(25k)
(6 cos 30°i - 6 sin 30°k)
{-130j} lb. in.
= r AB X
=
=
X
(25k)
Ans.
SOLUTION II (SCALAR ANALYSIS)
A
Although this problem is shown in three dimensions, the geometry
---......._ is simple enough to use the scalar equation M = Fd, where
Y d = 6 cos 30° = 5.196 in., Fig. 3- 32d. Hence, taking moments of the
forces about either point A or point B yields
x
M = Fd = 25 lb (5.196 in.) = 129.9 lb· in.
(d)
Fig. 3-32
Applying the right-hand rule, M acts in the - j direction. Thus,
M = {-130j}lb · in.
Ans.
3.6
-
EXAMPLE
MOMENT OF A COUPLE
3.1 3
-
Replace the two couples acting on the pipe assembly in Fig. 3-33a by a
resultant couple moment.
z 125N~
'
3 4
'~
~
~l
y
lSON
(b)
(a)
(c)
Fig. 3-33
SOLUTION (VECTOR ANALYSIS)
The couple moment Ml> developed by the forces at A and B, can
easily be determined from a scalar formulation.
M 1 = Fd = 150N(0.4m) = 60N · m
By the right-hand rule, M 1 acts in the +i direction, Fig. 3- 33b. H ence,
M1
=
{60i} N · m
Vector analysis will be used to determine M 2, caused by forces at C
and D. If moments are calculated about point D , Fig. 3- 33a,
M 2 = r oe x Fe, then
M i = roe X Fe = (0.3i) X [125(~)j - 125(~) k )
=
(0.3i) x [lOOj - 75k]
=
{22.5j + 30k} N · m
=
30(i x j) - 22.5(i x k)
Since M 1 and M i are free vectors, Fig. 3-33b, they may be moved to
some arbitrary point and added vectorially, Fig. 3- 33c. The resultant
couple moment becomes
MR = M1
+ Mi
=
{60i + 22.5j + 30k} N · m
Ans.
11 5
116
CHAPT ER
3
FORC E SYST EM RESU LTANTS
FUNDAMENTAL PROBLEMS
F3-19. Determine the resultant couple moment acting on
the beam.
400N
F3-22. Determine the couple moment acting on the beam.
lOkN
400N
'
4m
IA
200N
.
B
[o.2n1
-~
-
lm
200N
2m-
3 n1
_J
-
•
300N
300N
Prob.F3-19
F3-20. Determine the resultant couple moment acting on
the plate.
4
[(
10 kN
Prob. F3-22
F3-23. Determine the resultant couple moment acting on
the pipe assembly.
2001b
(Mch = 300 lb·ft
y
.\'.
tM
c)2
= 250 lb·ft
· - - - - -4 ft - - - - - 1
3001b
300 1b
Prob. F3-23
F3-24. Determine the couple moment acting on the pipe
assembly and express the result as a Cartesian vector.
Prob.F3-20
FA= 450N
z
F3-21. Determine the magnitude of F so that the resultant
couple moment acting on the beam is 1.5 kN · m clockwise.
F
0.4m
~
B
2kN
x
-F
Prob. F3-21
)'
c
Prob. F3-24
3.6
117
MOMENT OF A COUPLE
PROBLEMS
3-54. A clockwise couple M = 5 N · m is resisted by the
shaft of the electric motor. Determine the magnitude of the
reactive forces -R and R which act at supports A and B so
that the resultant of the two couples is zero.
*3- 56. If the resultant couple of the three couples acting
on the triangular block is to be zero, determine the
magnitude of forces F and P.
z
l_F
c
150mm
j
-R
x
R
Prob. 3-56
Prob.3-54
3-55. A twist of 4 N · m is applied to the handle of the
screwdriver. Resolve this couple moment into a pair of couple
forces F exerted on the handle and P exerted on the blade.
3-57. If F = 125 lb, determine the resultant couple
moment.
3-58. Determine the magnitude of F so that the resultant
couple moment is 450 lb· ft, counterclockwise. Where on
the beam does the resultant couple moment act?
200lb
F
4N·m
~j
1.5 ft
200 lb
- - -2ft --....i
Smm
1
Prob. 3-55
-F
Probs. 3-57/58
118
CHAPTER
3
FORCE SYSTEM RESULTANTS
3-59. Determine the magnitude and coordinate direction
angles of the resultant couple moment.
3-61. Determine the resultant couple moment of the two
couples that act on the assembly. Specify its magnitude and
coordinate direction angles.
z
M 1 =40 1b ·ft
'()..
M 2 = 30 lb· ft
80 Jb
~~
- -y
x
x
Prob. 3-59
3
in. - -- 1 60 lb
Prob. 3-61
*3-60. Determine the required magnitude of the couple
moments M 2 and M 3 so that the resultant couple moment
is zero.
3-62. Express the moment of the couple acting on the
frame in Cartesian vector form. The forces are applied
perpendicular to the frame. What is the magnitude of the
couple moment? Take F = 50 N.
3-63. In order to turn over the frame, a couple moment is
applied as sihown. If the component of this couple moment
along the x axis is M.. = {- 20i) N · m, determine the
magnitude F of the couple forces.
z
0
y
F
L:~
/~
· ~
/
M1 =300N · m
Prob. 3-60
x
1.Sm_-/
-F
Probs. 3-62/63
3.6
*3--04. Express the moment of the couple acting on the
pipe in Cartesian vector form. What is the magnitude of the
couple moment? Take F = 125 N.
119
M OMENT OF A COUPLE
*3-68. Express the moment of the couple acting on the
rod in Cartesian vector form. What is the magnitude of the
couple moment?
3-65. If the couple moment acting on the pipe has a
magnitude of 300 N · m. determine the magnitude of the
forces applied to the wrenches.
z
-F = {4i - 3j
I
+ 4kl k
-F
600mm
y
A
F=
I- 4i + 3j
- 4kl kN
Prob. 3-68
F
Probs. 3-64/65
3-69. If F, = 100 N, Fi = 120 N. and Fj = 80 N,
determine the magnitude and coordinate direction angles
of the resultant couple moment.
3-66. If F = 80 N, determine the magnitude and
coordinate direction angles of the couple moment. The pipe
assembly lies in the x-y plane.
3-70. Determine the required magnitude of F 1• F2 , and
F3 so that the resultant couple moment is
~1c)R = (50i - 45j - 20k)
· m.
3-67. If the magnitude of the couple moment acting on the
pipe assembly is 50 N · m. determine the magnitude of the
couple forces applied to each wrench. The pipe assembly
lies in the x- y plane.
F• = (150 k) N
x
-F2 0.2 m
Probs. 3-66/67
Probs. 3-69no
120
CHAPTER
3
FORCE SYSTEM RESULTANTS
3. 7
SIMPLIFICATION OF A FORCE AND
COUPLE SYSTEM
Sometimes it is convenient to reduce a system of forces and couple
moments acting on a body to a simpler form by replacing it with an
equivalent system , consisting of a single resultant force and a resultant
couple moment. A system is equivalent if the external effects it produces
on a body are the same as those caused by the original force and couple
moment system. If the body is free to move, then the external effects of a
system refer to the translating and rotating motion of the body, or if the
body is fully supported, they refer to the reactive forces at the supports.
For example, consider holding the stick in Fig. 3-34a, which is subjected
to the force F at point A. If we attach a pair of equal but opposite forces
F and - F at point B, which is on the line of action of F, Fig. 3- 34b, we
observe that -F at B and F at A will cancel each other, leaving only F
at B, Fig. 3- 34c. Force F has now been moved from A to B without
modifying its external effects on the stick; i.e., the reaction at the grip
remains the same. This demonstrates the principle of transm issibility,
which states that a force acting on a body (stick) is a sliding vector since
it can be applied at any point along its line of action.
We can also use the above procedure to move a force to a point that is
not on the line of action of the force. If F is applied perpendicular to the
stick, as in Fig. 3- 35a, then we can attach a pair of equal but opposite forces
F and -F to B, Fig. 3- 35b. Force Fis now applied at B, and the other two
forces, Fat A and -F at B, form a couple that produces the couple moment
M = Fd, Fig. 3- 35c. Therefore, the force F can be moved from A to B
provided a couple moment M is added to maintain an equivalent system.
This couple moment is determined by taking the moment of F about B.
Since Mis actually a free vector, it can act at any point on the stick. In each
case in Fig. 3-35 the systems are equivalent. This causes a downward force
F and clockwise couple moment M = Fd to be felt at the grip.
(c)
Fig. 3-34
(a)
(b)
Fig. 3-35
(c)
3.7
SIMPLIFICATION OF A FORCE ANO COUPLE SYSTEM
System of Forces and Couple Moments. Using this method,
a system of several forces and couple moments acting on a body can be
reduced to an equivalent single resultant force acting at a point 0 and a
resultant couple moment. For example, in Fig. 3-36a, 0 is not on the line of
action of Fi, and so this force can be moved to point 0 provided a couple (a)
moment (M 0 ) 1 = r 1 X F is added to the body. Similarly, the couple
moment (Moh = r2 X F2 should be added to the body when we move F2
to point 0. Finally, since the couple moment M is a free vector, it can just
be moved to point 0. Doing this, we obtain the equivalent system shown in
Fig. 3-36b, which produces the same external effects on the body as that of
the force and couple system shown in Fig. 3-36a. If we sum the forces and
couple moments, we obtain the resultant force FR = F1 + F2 and the
resultant couple moment (M R)o = M + (M 0 ) 1 + (M 0 )i, Fig. 3-36c.
Notice that FR is independent of the location of point 0 since it is
simply a summation of the forces. H owever, (MR)o depends upon this
location since the moments M 1 and M 2 are determined using the position (b)
vectors r1 and r2, which extend from 0 to each force. Also note that
(MR)o is a free vector and can act at any point on the body, although
point 0 is generally chosen as its point of application.
We can now generalize the above method of reducing a force and
couple system to an equivalent resultant force FR acting at point 0 and a
resultant couple moment (MR)o by using the following two equations.
FR= l F
(MR)o = l Mo + l M
II
II
(3-17)
(c)
The first equation states that the resultant force of the system is
equivalent to the sum of all the forces; and the second equation states
that the resultant couple moment of the system is equivalent to the sum
of all the couple moments l M plus the moments of all the forces about
point 0 , I M0 . If the force system lies in the x- y plane and any couple
moments are perpendicular to this plane, then the above equations
reduce to the following three scalar equations.
(FR)x = 2.Fx
(FR)y = 2.Fy
(MR)o = lM0 + IM
(3-18)
Here the resultant force is determined from the vector sum of its two
components (FR)x and (FR)y-
Fig. 3-36
1 21
122
CHAPTER
3
FORCE S YSTEM RESULTANTS
The weights of these traffic lights can be replaced by their equivalent resultant force
WR = Wi + Wi and a couple mome nt (MR)o = \.\'id 1 + W2 d 2 at the suppor t, 0.
In both cases the support must provide the same resistance to translation and rotation
in order to keep the member in the horizontal position.
PROCEDURE FOR ANALYSIS
The following points should be kept in mind when simplifying a force
and couple moment system to an equivalent resultant force and
couple system.
• Establish the coordinate axes with the ongm located at
point 0 and the axes having a selected orientation.
Force Summation.
• If the force system is coplanar, resolve each force into its x and
y components. If a component is directed along the positive x
or y axis, it represents a positive scalar; whereas if it is directed
along the negative x or y axis, it is a negative scalar.
• In three dimensions, represent each force as a Cartesian vector
before summing the forces.
Moment Summation.
• When determining the moments of a coplanar force system
about point 0 , it is generally advantageous to use the principle
of moments, i.e., determine the moments of the components of
each force, rather than the moment of the force itself.
• In three dimensions use the vector cross product to determine
the moment of each force about point 0. Here the position
vectors extend from 0 to any point on the line of action of
each force.
3.7
I
EXAMPLE
SIM PUFlCATION OF A FORCE AND COUPLE SYSTEM
3.14
Replace the force and couple system shown in Fig. 3- 37a by an
equivalent resultant force and couple moment acting at point 0.
y
(3 kN)sin 30"
-,-
l==::::::;:::;===o~-j__
0.1 m
. ·.
F===~~---i~
O
-,-
--+---+o .
O.lm
0.1 m
-'- l=======li====·=· ==~:...
-
0.2m
- -0.3m- - I
-
0.2m
~ (5kN) <--~
SkN
5
4 kN
4 kN
(b)
(a)
SOLUTION
Force Summation. The 3 kN and 5 kN forces are resolved into
their x and y components as shown in Fig. 3- 37b. We have
..t (FR)x =
2F,;
(FR)x
=
(3 kN) cos 30° + ( ~) (5 kN)
=
5.598 kN ~
(FR)y = (3 kN) sin 30° - ( ~) (5 kN) - 4 kN
+ j (FR)y = IFy;
=
-6.50 kN
Using the Pythagorean theorem, Fig. 3- 37c, the magnitude of FR is
FR = V(FR)/ + (FR)/ = V (5.598 kN) 2 + (6.50 kN) 2 = 8.58 kN Ans.
I ts direction (} is
(} =
tan
_1
((FR) y)
(FR)x
=
tan
Moment Summation.
c+ (MR)o
(MR)o
=
=
-t(
6.50 kN )
5.598 kN
0
=
49.3
2Mo;
-2.46kN·m = 2.46kN·m )
This clockwise moment is shown in Fig. 3- 37c.
(M R)o = 2.46 kN ·m
...
0
FR
(FR)y = 6.50 kN
(t) (5 kN) (0.1 m)
- ( ~ ) (5kN)(0.5m)-(4kN)(0.2m)
=
6.50 kN !
Ans.
Referring to Fig. 3- 37b, we have
(3 kN) sin 30°(0.2 m) - (3 kN) cos 30°(0.1 m) +
=
Ans.
NOTE: Realize that the resultant force and couple moment in Fig. 3- 37c
will produce the same external effects or reactions at the wall support as
those produced by the force system, Fig. 3- 37a.
(c)
Fig. 3-37
123
124
I
CHAPTER
EXAMPLE
3
FORCE SYSTEM RESULTANTS
3.1 s
I
Replace the force and couple system acting on the member in Fig. 3-38a
by an equivalent resultant force and couple moment acting at point 0.
SOON
y
750N
(MR)o = 37.5 N·m
l
0
•
1.25m+
200N
0
0
I
x
lm
l.25m -
11
(FR)x = 300 N
J
200N
(FR)y = 350 N
(a)
FR
(b)
Fig. 3-38
SOLUTION
Since the couple forces of 200 N are equal but
opposite, they produce a zero resultant force, and so it is not necessary
to consider them in the force summation. The SOO-N force is resolved
into its x and y components, thus,
Force Summation.
~(FR)x = IF,,; (FR)x = (~)(SOON) = 300N ~
+ j(FR)y = IFy; (FR)y = (SOON)(~) -7SON = -3SON = 3SON!
From Fig. 3- lSb, the magnitude of FR is
FR
=
V(FR).i + (FR)}
=
Y (300 N) 2 + (3SO N)2
= 461 N
Ans.
And the angle () is
() =
tan- 1 (~~:~:) = tan-1 (~~~~) = 49.4°
Ans.
Since the couple moment is a free vector, it can
act at any point on the member. Referring to Fig. 3-38a, we have
Moment Summation.
C+ (MR)o = IM0 + IM
(MR)o = (SOON) (~) (2.S m) - (SOON) (~) (1 m)
- (7SO N)(l.2S m) + 200 N · m
= -37.SN·m = 37.SN · m)
This clockwise moment is shown in Fig. 3- 38b.
Ans.
3.7
-
EXAMPLE
125
SIM PUFlCATION OF A FORCE AND COUPLE SYSTEM
3.16
-
z
The member is subjected to a couple moment M and forces Fi and F2
in Fig. 3- 39a. Replace this system by an equivalent resultant force and
couple moment acting at its base, point 0.
SOLUTION (VECTOR ANALYSIS)
The three-dimensional aspects of the problem can be simplified by using
a Cartesian vector analysis. Expressing the forces and couple moment as
Cartesian vectors, we have
lm
Fi
=
{-800k} N
F2
=
(300 N)u ca
=
(300 N)(rca)
y
rca
M
re
{-0.15i + O.lJ.} m
=
300 N [
=
-500 (~)j + 500(~ ) k
V (-0.15 m) 2 + (0.1 m)2
=
J
=
(a)
{-249.6i + 166.4j} N
z
{-400j + 300k}N · m
Force Summation.
(MR)
FR
+ F 2 = -800k - 249.6i + 166.4j
{-250i + 166j - 800k} N
=
Ans.
><io
x
Moment Summation.
=
M
+ re
X Fi
FR
(b)
(MR)o = 2 M + 'LMo
(MR) 0
0
= Fi
+ ra
Fig. 3-39
X F2
i
(MR)o = (-400j + 300k) + (l k)
X
(-800k) +
-0.15
-249.6
=
(-400j + 300k) + (0) + (-166.4i - 249.6j )
=
{-166i - 650j + 300k} N · m
The results are shown in Fig. 3- 39b.
j
0.1
166.4
k
1
0
Ans.
y
126
CHAPT ER
3
FORC E SYST EM RESU LTANTS
PRELIMINARY PROBLEM
P3-5. In each case, determine the x and y components of the
resultant force and the resultant couple moment at point 0.
SOON
!
0
SOON 400N
0
"r,t;r~----
lOON
r - -2 m - l - 2 m -
;f i
-+-1
l""-------'-1----o,-I_,. 200 N
l - 2m - - - ! - 2m - - 2m - I
SOON
(a)
(c)
l - 2m - I
SOON
SOON
2m
SOON
300N
0
i
4
)
2m ~
2m
200N·m
2m
0
I
(b)
(d)
Prob. P3-5
3.7
127
SIMPUFlCATION OF A FORCE AND COUPLE SYSTEM
FUNDAMENTAL PROBLEMS
F3-25. Replace the loading by an equivalent resultant
force and couple moment acting at point A.
F3-28. Replace the loading by an equivalent resultant
force and couple moment acting at point A.
. » lOO lb
100 lb
I
1 ft
A
_ . _ SO lb
..
AtI====t-==~=::=~~ . J L200 lb
Prob.F3-28
Replace the loading by an equivalent resultant
force and couple moment acting at point 0.
F3-29.
1 - -3 ft - -- - -3 ft - I
z
150 lb
h~
Prob. F3-25
F1 = {- 300i
~ 2n~~.5m~B
F3-26. Replace the loading by an equivalent resultant
force and couple moment acting at point A.
0
40N
+ 150j + 200k} N
~l~
.
x
200N·m
A•
~3m---
Prob. F3-29
SON
Replace the loading by an equivalent resultant
force and couple moment acting at point 0.
F3-30.
Prob. F3-26
z
Replace the loading by an equivalent resultant
force and couple moment acting at point A.
F3-27.
- 0-.7-5 _m _1·-0-.7-5-m-1 0.75 m
Prob. F3-27
a.1sm
F 1 = lOON
Mc = 75N·m
I
y
Prob.F3-30
128
CHAPTER
3
FORCE SYSTEM RESULTANTS
PROBLEMS
3-71. Replace the force system by an equivalent resultant
force and couple moment at point 0.
3-75. Replace the loading acting on the beam by an
equivalent resultant force and couple moment at point 0.
*3-72. Replace the force system by an equivalent resultant
force and couple moment at point P.
)'
)'
450N
455N
~
-r
1 - - - 2.5 m _ ___,
0.2 m
0
--+--+------~---~~~--- x
T\
200 N · m
o~
l ~~~~-x
2m
o.1sm I
/
0.75m
~ 1.5 m ~ 2 m ___',,____ 1.5 m - 1
·;;r
200N
lm-lp
Prob. 3-75
600N
Probs. 3-71172
3-73. Replace the loading acting on the beam by an
equivalent force and couple moment at point A.
3-74. Replace the loading acting on the beam by an
equivalent force and couple moment at point B.
*3-76. Replace the loading acting on the post by an
equivalent resultant force and couple moment at point A.
3-77. Replace the loading acting on the post by an
equivalent resultant force and couple moment at point B.
3kN
650N
2.5 kN
l- 2m -
4m
Probs. 3-73/74
Q
- 2m - I
SOON
300N
1500N ·m i A
~3m-1
60° B
sm- l -2m -I
Probs. 3-76177
3.7
3-78. Replace the loading acting on the post by a resultant
force and couple moment at point 0.
SIMPLIFICATION OF A FORCE AND C OUPLE SYSTEM
129
*3-80. ·me forces F1 = {- 4i + 2j - 3kl kN and F 2 =
(3i - 4j - 2kl kN act on the end of the beam. Replace
these forces by an equivalent force and couple moment
acting at point 0.
3001b
2 ft
-+l
2 ft
oo
I
_J_
2 ft
0
y
2001b
x
Prob. 3-80
Prob. 3-78
3-79. Replace the loading acting on the frame by an
equivalent resultant force and couple moment acting at
point A.
3-81. A biomechanical model of the lumbar region of the
human trunk is shown. The forces acting in the four muscle
groups consist of FR = 35 N for the rectus, F0 = 45 N for
the oblique. FL = 23 N for the lumbar latissimus dorsi, and
F£ = 32 N for the erector spinae. These loadings are
symmetric with respect to the y-z plane. Replace this system
of parallel forces by an equivalent force and couple moment
acting at the spine, point 0. Express the results in Cartesia n
vector form.
A
300N
05m
lm
SOON
....
f - o.5 m
- -0.3 m-1
400N
Prob. 3-79
y
x
Prob. 3-81
1 30
CHAPTER
3
FORCE SYSTEM RESULTANTS
3-82. Replace the loading by an equivalent resultant
force and couple moment at point 0. Take
F3 = {- 200i + 500j - 300k} N.
*3-84. Replace the force of F = 80 N acting on the pipe
assembly by an equivalent resultant force and couple
moment at point A.
z
z
F 1 =300N
x
x
Prob. 3-84
Prob. 3-82
3-83. Replace the loading by an equivalent resultant force
and couple moment at point 0.
z
3-85. The belt passing over the pulley is subjected to forces
F1 and Fi , each having a magnitude of 40 N. F1 acts in the -k
direction. Replace these forces by an equivalent force and
couple moment at point A. Express the result in Cartesian
vector form. Set e = 00 so that Fi acts in the - j direction.
3-86. The belt passing over the pulley is subjected to two
forces F1 and Fi , each having a magnitude of 40 N. F1 acts in
the -k direction. Replace these forces by an equivalent
force and couple moment at point A. Express the result in
Cartesian vector form. Take e = 45°.
z
0
x ~------rr~
O.Sm
)'
-1
0.7m
F2 = l- 2 i + 5 j - 3 k} kN
- - - 0.8 m - - -1
F1 = 18 i - 2 k} kN
Prob. 3-83
Probs. 3-85/86
3.8
3.8
FURTHER SIMPLIFICATION OF A FORCE ANO COUPLE SYSTEM
FURTHER SIMPLIFICATION OF
A FORCE AND COUPLE SYSTEM
In the preceding section, we developed a way to reduce a force and
couple moment system acting on a rigid body into an equivalent resultant
force FR acting at a specific point 0 and a resultant couple moment
(M R)o. The force system can be further re duced to an equivalent single
resultant force provided the lines of action of FR and (MR)o are
perpendicular to each other. This occurs when the force system is either
concurrent, coplanar, or parallel.
Concurrent Fo ·ce Sy ·tem. Since a concurrent force system is one
Y
in which the lines of action of all the forces intersect at a common point 0 ,
Fig. 3-40a, then the force system produces no moment about this point.
As a result, the equivale nt system can be represented by a single resultant
force FR = I F acting at 0, Fig. 3-40b.
(a)
Coplanar Force System. In the case of a coplanar force system, the
II
lines of action of all the forces lie in the same plane, Fig. 3-41a, and so the
resultant force FR = I F of this system also lies in this plane. Furthermore,
the moment of each of the forces about any point 0 is directed perpendicular
to this plane. Thus, the resu ltant moment (MR)o and resultant force FR will
be mULually perpendicular, Fig. 3-4lb. The resultant moment can be
replaced by moving the resultant force FR a perpendicular or moment arm
distance d away from point 0 such that FR produces the same moment
(MR)o about point 0, Fig. 3-4lc. This distanced can be determined from
the scalar equation (MR)o = F~ = "2.Mo or d = (MR)o/ FR.
(b)
Fig. 3-40
(MR)o
ov
_.,g__
F1
F4
(a)
(b)
Fig. 3-41
(c)
FR
1 31
1 32
CHAPTER
3
FORCE SYSTEM RESULTANTS
Parallel Force System. The parallel force system shown in
Fig. 3-42a consists of forces that are all parallel to the z axis. Thus, the
resultant force FR = IF at point 0 must also be parallel to this axis,
Fig. 3-42b. The moment produced by each force lies in the plane of the
plate, and so the resultant couple moment, (MR)o, will also lie in this
plane, along the moment axis a since FR and (MR)o are mutually
perpendicular. As a result, the force system can be further reduced to an
equivalent single resultant force FR, acting through point P located on
the perpendicular b axis, Fig. 3-42c. The distance d along this axis from
point 0 requires (MR)o = FRd = lM0 or d = lM0 / FR.
PROCEDURE FOR ANALYSIS
0
The technique used to reduce a coplanar or parallel force system to
a single resultant force follows a similar procedure outlined in the
previous section.
• Establish the x, y, z axes and locate the resultant force FR an
arbitrary distance away from the origin of the coordinates.
Force Summation.
• The resultant force is equal to the sum of all the forces in the
system.
• For a coplanar force system, resolve each force into its x and y
components. Positive components are directed along the
positive x and y axes, and negative components are directed
along the negative x and y axes.
The four cable forces are all concurrent at
point 0 on this bridge tower. Consequently
they produce no resultant moment there, only
a resultant force FR· Note that the designers
have positioned the cables so that FR is
directed along the bridge tower directly to the
support, so that it does not cause any bending
of the tower.
Moment Summation.
• The moment of the resultant force about point 0 is equal to
the sum of all the couple moments in the system plus the
moments of all the forces in the system about 0, lM0 .
• To find the location d of the resultant force from point 0, use
the condition that d = lM0 / FR.
3.8
EXAMPLE
FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
3 .17
Replace the force and couple moment system acting on the beam in
Fig. 3-43a by an equivalent resultant force, and find where its line of
action intersects the beam. measured from point 0.
y
x
-1.5m
(b)
(a)
Fig. 3-43
SOLUTION
Force Summation. Summing the force components,
..:!+ (FR)x = IF,.;
(FR)x = 8 kN(~) = 4.80 kN ~
+f(FR)y =IF,,;
(FR),, = -4 kN + 8 kN{~) = 2.40 kNf
From Fig. 3-44b, the magnitude of FR is
FR
= V(4.80 kN) 2 + (2.40 kN) 2 = 5.37 kN
Ans.
The angle 8 is
- 660
8 -- tan - ·(2.40kN)
4.80 kN - 2 ·
An~
Moment Summation. We must equate the moment of FR about
point 0 in Fig. 3-43b to the sum of the moments of the force and
couple moment system about point 0 in Fig. 3-43a. Since the line of
action of (FR)., passes through point 0, only (FR)y produces a moment
about this point. Thus,
C+ (MR)o =
IM0 ;
2.40 kN(d)
= -(4 kN)(l.5 m)
- 15 kN · m
-[8kN(~)j(0.5m) + [8 kN(~) j(4.5m)
d
= 2.25 m
Ans.
133
1 34
I
3
CHAPTER
FORCE S YSTEM RESULTANTS
3.1 s
EXAMPLE
I
y
3"""
tt_ 1_ _"""
s-'-'
ft~-I
1""""""
3 ft
The jib crane shown in Fig. 3--44a is subjected to three coplanar forces.
Replace this loading by an equivalent resultant force and specify where
the resultant's line of action intersects the column AB and boom BC.
B
I
SOLUTION
6 ft
Resolving the 250-lb force into x and y
components and summing the iforce components yields
Force Summation.
60 Jb
175 lb
~(FR), = 2F,; (FR), = -250lb ( ~) -175lb
5 ft
-325lb
=
=
325lb ~
+ j(FR), = 2£,.; (FR), = -250lb ( ~) -60lb = -260lb = 260lb!
A
(a)
As shown by the vector addition in Fig. 3-44b,
FR =
y
8
V (325 lb) 2 + (260 lb) 2 =
=
tan
- 1(260 lb)
325 lb
=
38
0
416 lb
.7 7
Ans.
Ans.
B
/
FR
/
/
/
/
/
/
/
260 Jb
Moments will be summed about point A.
Assuming the line of action of lFR intersects AB at a distance y from A ,
Fig. 3-44b, we have
Moment Summation.
/
/
/
/
~
+ (MR)A = IMA;
=
325 lb (y) + 260 lb (0)
175 lb(5ft) - 60lb(3ft) + 250lb ( ~ ) (llft) - 250lb ( ~)(8ft)
y = 2.29 ft
(b)
Ans.
Fig. 3-44
By the principle of transmissibility, FRcan also be placed at a distance x
where it intersects BC, Fig. 3-44b. In this case we have
~
+ (MR)A = IMA; 325 lb (11 ft) - 260 lb (x)
=
175lb(5ft) - 60lb(3ft) + 250lb ( ~ ) (llft) - 250lb ( ~)(8ft)
x = 10.9 ft
Ans.
3.8
EXAMPLE
135
FURTHER SIMPUFlCATION OF A FORCE AND COUPLE SYSTEM
3.19
-
-
The slab in Fig. 3-45a is subjected to four parallel forces. Determine the
magnitude and direction of a resultant force equivalent to the given force
system, and locate its point of application on the slab.
z
400N
600N
+
0
y
P(x, y ) • - - L -
x
(a)
x
(b}
Fig. 3-45
SOLUTION (SCALAR ANALYSIS)
Force Summation.
+ !FR = IF;
From Fig. 3-45a, the resultant force is
FR = 600N - lOON + 400N + SOON
= 1400 N!
Ans.
We require the moment about the x axis of
the resultant force, Fig. 3-45b, to be equal to the sum of the moments
about the x axis of all the forces in the system, Fig. 3-45a. The moment
arms are determined from they coordinates, since these coordinates
represent the perpendicular distances from the x axis to the lines of
action of the forces. Using the right-hand rule, we have
Moment Summation.
(MR)x
=
lMx;
-(1400 N)y
=
600 N(O) + 100 N(S m) - 400 N(lO m) + 500 N(O)
-1400y = -3500
y = 2.50m
Ans.
In a similar manner, a moment equation can be written about the
y axis using moment arms defined by the x coordinates of each force.
(MR) y = lMy;
(1400 N)x
1400x
=
=
600 N(8 m) - 100 N(6 m) + 400 N(O) + 500 N(O)
4200
x = 3m
Ans.
FR = 1400 N placed at point P(3.00 m, 2.50 m) on
the slab, Fig. 3-45b, is therefore equivalent to the parallel force system
acting on the slab in Fig. 3-45a.
NOTE: A force of
x
1 36
I
CHAPTER
EXAMPLE
3
3.20
FORCE SYSTEM RESULTANTS
I
Fs = 500 lb
2
Replace the force system in Fig. 3-46a by an equivalent resultant
force and specify its point of application on the pedestal.
in. SOLUTION
Force Summation.
Summing forces,
FR
=
IF; FR
y
(a)
Here we will demonstrate a vector analysis.
=
FA + Fs + Fe
=
{-300k} lb + {-500k} lb + {lOOk}lb
=
l-700k }lb
Ans.
Location. Moments will be summed about point 0. The resultant
force FR is assumed to act through point P (x, y, 0), Fig. 3-46b. Thus
(MR)o
2 Mo;
=
rp x FR
=
(rA x FA)
+
(r8 x F8 )
+ (r e x Fe)
(xi + yj ) x (-700k ) = ((4i) x (-300k)]
+ ((-4i + 2j) x (-500k)J + ((-4j) x (lOOk)]
-700x(i
x
X
k) - 700y(j
X
k) = -1200(i
X
k) + 2000(i
X
k)
- lOOO(j x k) - 400(.j x k)
700xj - 700yi
(b)
120Qj - 2000j - lOOOi - 400i
=
Equating the i and j components,
-700y
Fig. 3-46
=
-1400
y = 2 in.
700x
=
(1)
Ans.
-800
x = -1.14 in.
(2)
Ans.
The negative sign indicates that the x coordinate of point P is
negative.
NOTE: As demonstrated in Example 3.19, it is also possible to
establish Eq. 1 and 2 directly by summing moments about the x and
y axes. Using the right-hand ruk, we have
2Mr;
-700y
=
-100 lb(4 in.) - 500 lb(2 in.)
(MR) y = 2My;
700x
=
300 lb(4 in.) - 500 lb(4 in.)
(MR)x
=
3.8
1 37
FURTHER SIMPLIFICATION OF A FORCE ANO COUPLE SYSTEM
PRELIMINARY PROBLEMS
P3-6. In each case. determine the x and y components of
the resultant force and specify the distance where this force
acts from point 0.
1'3-7. In each case. determine the resultant force and
specify its coordinates x and y where it acts on the x-y plane.
200N
200N
100 N
260N
200N
0
•
l-2m--2111--2m- I
(a)
(a)
200N
100 N
SOON
400N
y
f
~o
i
I
l-2m
I
2m~
x
(b)
(b)
200N
IOON
400N
500 N
O
5
300N
SOON
;t
~
2m
?'
600N· m
0
1
)
l-2m-~2m -l-2m -I
(c)
(c)
Prob. P3-6
Prob. P3-7
1 38
CHAPTER
3
FORCE SYSTEM RESULTANTS
FUNDAMENTAL PROBLEMS
F3-3L Replace the loading by an equivalent resultant
force and specify where the resultant's line of action
intersects the beam, measured from 0.
F3-34. Replace the loading by an equivalent resultant
force and specify where the resultant's line of action
intersects the member AB, measured from A.
y
O.Sm
- 1.Sm-1
y
5001b
•
250 lb
5001b
0.5m
0.5m
•
0
·1
'
1-3ft--3ft--3ft--3ft-I
B
8 kN
x
3m
Prob.F3-31
F3-32. Replace the loading by an equivalent resultant
force and specify where the resultant's line of action
intersects the member, measured from A.
F3-35. Replace the loading by an equivalent single
resultant force and specify the x and y coordinates of its line
of action.
2001b
1-3ft
Prob.F3-34
3ft1-3ftl 3(f
z
501b
400N
A
HOON
lOOlb
Prob.F3-32
x
F3-33. Replace the loading by an equivalent resultant
force and specify where the resultant's line of action
intersects the horizontal segment of the member, measured
from A.
Prob.F3-35
F3-36. Replace the loading by an equivalent single
resultant force and specify the x and y coordinates of its line
of action.
z
20kN
2m
2m -
2m
Prob.F3-33
x
Prob.F3-36
3.8
139
FURTHER SIMPLIFICATION OF A FORCE ANO COUPLE SYSTEM
PROBLEMS
3-87. The weights of the various components of the truck
are shown. Replace this system of forces by an equivalent
resultant force and specify its location, measured from B.
*3-88. The weights of the various components of the truck
are shown. Replace this system of forces by an equivalent
resultant force and specify its location, measured from
point A.
3-91. Replace the loading by a single resultant force.
Specify where the force acts. measured from end A.
*3-92. Replace the loading by a single resultant force.
Specify where the force acts. measured from B.
700N
300N
~2 m-1
---4 111---'
Probs. 3-91/92
3-93. Replace the loading by a single resultant force.
Specify where its line of action intersects a vertical line
along member AB. measured from A.
Probs. 3-87188
3-89. Replace the three forces acting on the shaft by a
single resultant force. Specify where the force acts., measured
from end A.
3-90. Replace the three forces acting on the shaft by a
single resultant force. Specify where the force acts., measured
from end B.
1- - -s
400N
200N
B
200N
!-O.S m--0.Sm-!
(i()() N
-- o
0
1.5 m
ri--1-3 n- 1-2 n-1- 4
r1 - - 1
B
A
5001b
A
200lb
Pro bs. 3-89/90
260lb
Pro b. 3-93
c
140
CHAPTER
3
FORCE SYSTEM RESULTANTS
3-94. Replace the loading on the frame by a single
resultant force. Specify where its line of action intersects a
vertical line along member AB, measured from A.
3-98. Replace the parallel force system acting on the plate
by a resultant force and specify its location on the
x-z plane.
3-95. Replace the loading on the frame by a single
resultant force. Specify where its line of action intersects a
horizontal line along member CB, measured from end C.
z
y
-
1 n1 ~1
600N
0.5 m
B
400N
1.5 m
l
900 N
1'-"'--~-400-N
--'I:'___.._
'
lm
x
Probs. 3-94195
y
~
3kN
0.5m
x
Prob. 3-98
*3-96. Replace the loading acting on the post by a
resultant force, and specify where its line of action intersects
the post AB, measured from point A.
3-97. Replace the loading acting on the post by a resultant
force, and specify where its line of action intersects the post
AB, measured from point B.
3-99. Replace the loading acting on the frame by an
equivalent resultant force and specify where the resultant's
line of action intersects member AB, measured from A.
A
B
I
lm
_J_
SOON
~
150 lb
lm
I
300N
lm
I
I
2 ft
A
50 ~----++-BIJ
/1 __3ft - - 1
~
50lb
Probs. 3-96197
4 ft
Prob. 3-99
3.8
FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
*3-100. Replace the loading acting on the frame by an
equivalent resultant force and specify where the resultant's
line of action intersects member BC, measured from B.
A
141
3-102. Determine the magnitudes of FAand F8 so that the
resultant force passes through point 0.
r
2 ft
_l
SkN
4
l50lb
lOOmm
4 ft
~.J
x
1---3r1-I
Prob. 3-102
Prob. 3-100
3-101. If FA = 7 kN and F8 = 5 kN. represent the force
system by a resultant force, and specify its location on the
x- y plane.
3-103. The tube supports the four parallel forces. Determine
the magnitudes of forces Fe and F0 acting at C and D so
that the equivalent resultant force of the force system acts
through the midpoint 0 of the tube.
600N
SkN
100mm
x
Prob.3-101
Prob. 3-103
142
CHAPTER
3
FORCE SYSTEM RESULTANTS
*3-104. The building slab is subjected to four parallel
column loadings. Determine the equivalent resultant force
and specify its location (x, y) on the slab. Take F 1 = 8 kN
andF2 = 9 kN.
3-106. If FA = 40 kN and F8 = 35 kN, determine the
magnitude of the resultant force and specify the location of
its point of application (x, y) on the slab.
z
30kN
12kN
0.75m
90kN
20kN
x
Prob. 3-106
Prob. 3-104
3-105. The building slab is subjected to four parallel
column loadings. Determine F1 and F2 if the resultant force
acts through point (12 m, 10 m).
3-107. If the resultant force is required to act at the center
of the slab, determine the magnitude of the column loadings
FA and F8 and the magnitude of the resultant force.
z
12kN
30kN
0.75 m
90kN
0.75m
Prob. 3-105
Prob.3-107
3.9
3.9
143
REDUCTION OF A SIMPLE DISTRIBUTED LOADING
REDUCTION OF A SIMPLE
DISTRIBUTED LOADING
p
Sometimes a body may be subjected to a loading that is distributed over
its surface. For example, wind on the face of a sign, water within a tank,
or sand on the floor of a storage container all exert distributed loa dings.
The pressure caused by a loading at each point on the surface represents
the intensity of the loading. It is measured using pascals, Pa (or N /m2 ) in
SI units, or lb/ft 2 in the U.S. Customary system.
x
(a)
Loading Along a Single Axis. The most common type of
distributed pressure loading is represented along a single axis. For
example, consider the beam (or plate) in Fig. 3-47a that has a constant
width and is subjected to a pressure loading that varies only along the
x axis. This loading can be described by the function p = p(x) N/m2 .
Since it contains only one variable, x, we can represent it as a coplanar
distributed load. To do so, we must multiply it by the width b m of the
beam, so that w(x) = p(x)b N/m, Fig. 3-47b. Using the methods of
Sec. 3-8, we can replace this coplanar parallel force system with a single
equivalent resultant force FR, Fig. 3-47c.
w
dF=dA
w = w(x)
I
0
dx- -
x-1
L
(b)
w
FR
c
Magnitude o
Re u tant Force. The magnitude of FR is
equivalent to the sum o f all the forces in the system, FR = 'i.F. In this
case integration must be used since there is an infinite number of parallel
forces dF acting on the beam. Fig. 3-47b. Each dF is acting on an element
of length dx, and since w(x) is a force per unit length, then
dF = w(x) dx = dA where dA is the colored differential area under the
loading curve. For the entire length L ,
(3-19)
Therefore, the magnilllde of the resultant force is equal to the area A. under
the loading diagram, Fig. 3-47c.
o x-l
L
(c)
Fig. 3-47
x
144
CHAPTER
3
FORCE SYSTEM RESULTANTS
p
Location of Resultant Force. The location
FR can be
determined by equating the moments of the force resultant and the
parallel force distribution about point 0 (the y axis), (MR)o = 2M0 .
Since dF produces a moment of x dF = xw(x) dx about 0, Fig. 3-47b,
then for the entire length L, Fig. 3-47c, we have
p = p (x)
-xFR
x
Solving for
[xw(x) dx
[xw(x) dx
(3- 20)
x =
IV
0
= -
x, using Eq. 3- 19, we have
(a)
dF=dA
w = w(x )
I
x of
dx - -
1---x--I
L- - -
(b)
IV
c
.:!__v-1
L ---
/ . w(x) dx
This coordinate x locates the geometric center or centroid of the area
under the distributed loading. In other words, the line of action of the
resultant force passes through the centroid C (geometric center) of the area
under the loading diagram, Fig. 3-47c. When the distributed-loading
diagram is in the shape of a rectangle, triangle, or some other simple
geometric form, then the centroid location for such common shapes does
not have to be determined from the above equation. Rather it can be
obtained directly from the tabulation given on the inside back cover.
Once xis determined, FR by symmetry passes through point (x, 0) on
the surface of the beam, Fig. 3-47a, and so in three dimensions the
resultant force has a magnitude equal to the volume under the loading
curve p = p(x) and a line of action which passes through the centroid
(geometric center) of this volume.
(c)
Fig. 3-47 (Repeated)
IMPORTANT POINTS
• Coplanar distributed loadings are defined by using a loading
function w = w(x) that indicates the intensity of the loading
along the length of a member. This intensity is measured in
N/m or lb/ft.
• The external effects caused by a coplanar distributed load acting
on a member can be represented by a resultant force.
The pile of brick creates an approximate
triangular distributed loading on the board.
• This resultant force is equivalent to the area under the loading
diagram, and has a line of action that passes through the centroid
or geometric center of this area.
3.9
EXAMPLE
145
REDUCTION OF A SIMPLE DISTRIBUTED LOADING
3 .21
Determine the magnitude and location of the equivalent resultant force
acting on the shaft in Fig. 3--4&.
w
w
240Nfm
w = (60r)N/m
dA = wdx
x
0
-
- x
0
x
2m
- dx
i=
I.S m-I
(b)
(a)
Fig. 3-48
SOLUTION
Since w
=
w(x) is given, this problem will be solved by integration.
The differential element has an area dA
= w dx = 60x 2 dx. Applying
Eq. 3- 19,
+ !FR = I.F;
= 160N
Ans.
The location x of FR me asured fro m 0 , Fig. 3-48b, is de te rmined from
Eq. 3- 20.
x =
=
160N
l.5m
Ans.
NOTE: These results can be c hecked by using the ta ble in A ppendix B,
where for the exparabolic area of length a, height b, and shape shown
in Fig. 3-48a, we have
A
ab
= -3 =
2 m(240 N/ m)
3
3
3
= 160 N and x = -a
= -(2
m) = 1.5 m
4
4
146
CHAPTER
I EX AMPLE
3
FORCE SYSTEM RESULTANTS
3.22
A distributed loading of p = (800x) Pa acts over the top surface of the
beam shown in Fig. 3-49a. Determine the magnitude and location of the
equivalent resultant force.
7200 Pa
p = 800x Pa
x
Y-.....__
IV
iv=
160x N/m
1440N/m
(a)
SOLUTION
- -x- - - i
- - - -9m- - - - I
(b)
Since the loading intensity is uniform along the width of the beam
(they axis), the loading can be viewed in two dimensions as shown in
Fig. 3-49b. Here
w = (800x N/m2)(0.2 m)
FR= 6.48 kN
1--x = 6 m---1--3 m-J
(c)
Fig. 3-49
=
(160x) N/m
At x = 9 m, w = 1440 N/m. We may again apply Eqs. 3-19 and 3-20 as
in the previous example; however, here it is easier to find the area and its
centroid using Appendix B.
The magnitude of the resultant force is equivalent to the area of the
triangle.
FR = !(9 m)(1440 N/m)
=
6480 N
=
6.48 kN
Ans.
The line of action of FR passes through the centroid C of this triangle.
Hence,
x=
9 m - l(9 m)
=
Ans.
6m
The results are shown in Fig. 3-49c.
FR as acting through the centroid
of the volume of the loading diagram p = p(x) in Fig. 3-49a. Then FR
intersects the x- y plane at the point (6 m, 0). Furthermore, the
magnitude of FR is equal to the volume under this loading diagram; i.e.,
NOTE: We can also view the resultant
FR = V
=
!(7200 N/m2 )(9 m)(0.2 m)
=
6.48 kN
Ans.
3.9
147
REDUCTION OF A SIMPLE DISTRIBUTED LOADING
3.23
EXAMPLE
The granular material exerts the distributed loading on the beam as
shown in Fig. 3-50a. D etermine the magnitude and location of the
equivalent resultant of this load.
100 lb/ ft
SOLUTION
A
1-"_ _ _ _ _.........,.,
50 lb/ fl
The
8
area of the loading diagram is a trapezoid, and therefore the
solution can be obtained directly from the formulas for a trapezoid
listed in Appendix B. Since these formulas are not easily remembered,
instead we will solve this problem by using "composite" areas. H e re
we will divide the loading into a rectangular and a triangular loading
as shown in Fig. 3- 50b. The magnitude of the force represented by
each of these loadings is equal to its associated area,
r-----9 f t - - - (a)
Fi
Fi
B
-
-'1'2 -
1- - - - 9 f l - - - -1
=
=
!(9 ft)(50 lb/ft) = 225 lb
(9 ft)(50 lb/ft) = 450 lb
The lines of action of these parallel forces act thro ugh the respective
centroids of their associated areas and therefore intersect the beam at
{b)
xi =
t<9 ft) = 3 ft
Xi = !(9 ft)
= 4.5 ft
The two parallel forces F 1 and F2 can be reduced to a single resultant
FR· The magnitude of FR is
B
FR= 225 + 450
(c)
I
j_
= 675 lb
Ans.
We can find the location o f FR with reference to point A , Fig. 3-50b
and Fig. 3-50c. We require
( + (MR)A
= ~MA;
= 3(225) +
x = 4 ft
.X(675)
100 lb/ ft
4.5(450)
Ans.
·,A
NOTE: The trapezoidal area in Fig. 3- 50a can a lso be divided into two
·----9 f l - - - -1
triangular areas as shown in Fig. 3- 50d. In this case
{d)
Fig. 3-50
~
=
F4
=
!(9 ft)(lOO lb/ft) = 450 lb
!(9 ft)(50 lb/ft) = 225 lb
and
x3 =
~(9 ft) = 3 ft
~ =
9ft - t<9 ft) =
6 ft
Using these results, show again that FR = 675 lb and x = 4 ft.
148
CHAPTER
3
FORCE SYSTEM RESULTANTS
FUNDAMENTAL PROBLEMS
F3-37. Determine the resultant force and specify where it
acts on the beam, measured from A.
F3-40. Determine the resultant force and specify where it
acts on the beam, measured from A.
9 kN/m
200 lb /ft
6kN/m
~
I
lll
A
! I I I I I 1t t t tI
3kN/m
'°'
I
I
8
1SO lb/ft
n
A
•
- -6
-
1.5m- L3m
t
500 lb
'
=
r~,B
ft--1~3 ft~-3 tt-1
l.5m j
Prob.F3-40
Prob.F3-37
F3-38. Determine the resultant force and specify where it
acts on the beam, measured from A.
F3-41. Determine the resultant force and specify where it
acts on the beam, measured from A.
6kN/m
150lb/ft
A)~ l l l l l l l l !A~
I
~8ft
6ft
I
Prob.F3-38
F3-39. Determine the resultant force and specify where it
acts on the beam, measured from A.
A•
B
- - - 4.Sm - - -
l.Sm -
Prob.F3-41
F3-42. Determine the resultant force and specify where it
acts on the beam, measured from A.
6kN/m
160N/m
.•.1 - - - - -4m- - - - - -1
. 'I-----------~
Prob.F3-39
Prob.F3-42
3.9
149
REDUCTION OF A SIMPLE D ISTRIBUTED LOADING
PROBLEMS
*3-108. Replace the loading by an equivalent resultant
force and couple moment acting at point 0.
50 lb/ft
1 - - -9ft - - -
0
1- - - 9 ft - --
3-111. Currently eighty-five percent of all neck injuries
are caused by rear-end car collisions. To alleviate this
problem, an automobile seat restraint has been developed
that provides additional pressure contact with the cranium.
During dynamic tests the distribution of load on the
cranium has been plotted and shown to be parabolic.
Determine the equivalent resultant force and its location,
measured from point A.
1
50 lb/ft
Prob. 3-108
3-109. Replace the distributed loading with an equivalent
resultant force, and specify its location on the beam,
measured from point 0.
IV
--.a.- w = 12(1 + 2i-2) lb/ft
3kN/m
0
x
1-3m--~15m-
Prob.3-111
*3-112. Replace the distributed loading by an equivalent
resultant force, and specify its location on the beam,
measured from the pin at A.
Prob. 3-109
3-110. Replace the loading by an equivalent resultant force
and specify its location on the beam, measured from A.
4kN/m
2kN/m
IV
5 kN/m
-
2kN /m
x
~) ~'
I
4m
Prob. 3-110
~-201-I
~ 3 n1
3 n1 - - ---1
Prob. 3-112
150
C H APT ER
3
FORC E SYST EM RESU LTANTS
3-113. Replace the distributed loading by an equivalent
resultant force and specify where its line of action intersects
a horizontal line along member AB, measured from A.
*3-116. Determine the equivalent resultant force and
couple moment at point 0.
3-114. Replace the distributed loading by an equivalent
resultant force and specify where its line of action intersects
a vertical line along member BC, measured from C.
JV
3kN/m
A
B
--
•
I
9kN/m
I
3 n1
I
1 - - - - - -3 n1 - - - - - -- 1
2kN/m
4m
Prob.3-116
c_~.. Probs. 3-113/114
3-117. Determine the magnitude of the equivalent
resultant force and its location, measured from point 0.
3-115. Determine the length b of the triangular load and
its position a on the beam so that the equivalent resultant
force is zero and the resultant couple moment is 8 kN · m
clockwise.
IV
- a - ·1 - - - - b - - - -
JV=
6kN/m
... F=============::::!:::::::!:::::!:::::!::::!::::::!::::::l
4 lb/ft
,/..--
(4
---- ----
+ 2G) lb/ft
8.90 lb /ft
. · f-----~~~~~~~~~~-r--1
~: A
..
I
()
Prob.3-llS
Prob.3-117
x
CHAPTER REVIEW
CHAPTER REVIEW
Moment of Force-Scalar Definition
Moment axis
A force produces a turning effect or
moment about a point 0 that does not lie
on the force's line of action. In scalar
form, the moment magnimde is the
product of the force and the moment arm
or perpendicular distance from point 0 to
the line of action of the force.
~
I
Mo= Fd
~Mo
0
F
The direction of the moment is defined
using the right-hand rule. M 0 always acts
along an axis perpendicular to the plane
containing F and d, and passes through
the point 0.
Principle of Moments
Rather than finding d, it is normally easier
to resolve the force into its x and y
components, determine the moment of
each component about the point, and
then sum the results. This is called the
principle of moments.
M 0 = Fd = F,y - F,.x
Moment of a Force- Vector Definition
Since three-dimensional geometry is
generally more difficult to visualize, the
vector cross product should be used to
determine the moment. Here M 0 = r X F,
where r is a position vector that extends
from point 0 to any point A , B, or Con
the line of action of F.
If the position vector r and force F are
expressed as Cartesian vectors, then the
cross product can be evaluated from the
expansion of a determinant.
Mo =
F = ro
rA X
X
F =
re X
F
c
x
M0
=r
X
i
F = rx
F,
k
r,
F_
-
151
152
CHAPTER
3
FORCE SYSTEM RESULTANTS
Moment about an Axis
If the moment of a force F is to be
determined about a specific axis a, then
for a scalar sol ution the moment arm, or
shortest distance d 0 from the line of action
of the force to the axis must be used. This
distance is perpendicular to both the axis
and the line of action of the force.
F
a
When the line of action of F intersects the
axis then the moment of F about the axis
is zero. Also, when the line of action of Fis
parallel to the axis, the moment of F about
the axis is zero.
In three dimensions, the scalar triple
product should be used. Here u0 is the unit
vector that specifies the direction of the axis,
and r is a position vector that is directed
from any point on the axis to any point on
the line of action of the force. If M 0 is
calculated as a negative scalar, then the
sense of direction of M 0 is opposite to u0 .
r
M0
=u
0 •
(r x F)
=
u., u.1
rx
~
r,.
lt0 ;.
r,
F
F,. F,
Axis of projection
a'
Couple Moment
A couple consists of two equal but
opposite forces that act a perpendicular
distanced apart. Couples tend to produce
a rotation without translation.
The magnitude of the couple moment is
M = Fd, and its direction is established
using the right-hand rule.
If the vector cross product is used to
determine the moment of a couple, then r
extends from any point on the line of
action of one of the forces to any point on
the line of action of the other force F that
is used in the cross product.
M = Fd
M
=rx
F
F~-F
CHAPTER REVIEW
Simplification or a Force and
Couple System
F,
Any system of forces and couples can be
redured to a single resultant force and
resultant couple moment acting at a point 0.
The resultant force is the sum of all the
forces in the system. FR = I F, and the
resultant couple moment is equal to the
sum of the couple moments and the
moments of all the forres about point 0.
(M R)o = I M + I M0 .
Further simplification to a single resultant
force is possible, provided the force system
is concurrent, coplanar. or parallel. To find
the location of the resultant force from
point 0. it is necessary to equate the
moment of the resultant force about the
point to the moment of the forces and
couples in the system about the same point.
Fn
Fn
b
b
a
Coplanar Distributed Loading
A simple distributed loading can be
represented by its resultant force, which is
equivalent to the area under the loading
curve. This resultant has a line of action
that passes through the cemroid or the
geometric center of the area under the
loading diagram.
w
w = w(x)
- - -L - - - '
b
A11t0
d= -
FR
a
153
154
CHAPTER
3
FORCE SYSTEM RESULTANTS
REVIEW PROBLEMS
R3-L The boom has a length of 30 ft, a weight of 800 lb,
and mass center at G. If the maximum moment that can be
developed by a motor at A is M = 20(103) lb· ft, determine
the maximum load W, having a mass center at G ', that can
be lifted.
R3-3. The hood of the automobile is supported by the
strut AB, which exerts a force of F = 24 lb on the hood.
Determine tihe moment of this force about the hinged axis y.
z
B
x
2 ft
Prob. R3-1
Prob. R3-3
R3-2. Replace the force F having a magnitude of F = 50 lb
and acting at point A by an equivalent force and couple
moment at point C.
z
*R3-4. Friction on the concrete surface creates a couple
moment of M 0 = 100 N · m on the blades of the trowel.
Determine the magnitude of the couple forces so that the
resultant couple moment on the trowel is zero. The forces
lie in a horizontal plane and act perpendicular to the handle
of the trowel
A
30 ft
F
/
x
x
20 ft
y
Prob. R3-2
Prob. R3-4
REVIEW PROBLEMS
R3-5. Replace the force and couple system by an
equivalent force and couple moment at point P.
15 5
R3-7. The building slab is subjected to four parallel
column loadings. Determine the equivalent resultant force
and specify its location (x, y) on the slab. Take Fi = 30 kN,
Fi = 40kN.
Prob. R3-7
RJ-6. Replace the force system acting on the frame by a
resultant force. and specify where its line of action intersects
member AB, measured from point A.
*R.3-8. Replace the distributed loading by an equivalent
resultant force, and specify its location on the beam.
measured from the pin at C.
-2.s r1 -tB-3 r1
--1
I
2 rt
3001b
200 1b
250 lb
L
.f.
~~.·:
:
Prob. R3-6
- - --15(!
1800
----1
Prob. R3-8
lb/ft
lSft-- - - 1
CHAPTER
4
(© YuryZap/Shutterstock)
It is important to be able to determine the forces in the cables used to support
this boom to ensure that it does not fail. In this chapter we will study how to apply
equilibrium methods to determine the forces acting on the supports of a rigid
body such as this.
EQUILIBRIUM OF A
RIGID BODY
CHAPTER OBJECTIV ES
•
To develop the equations of equilibrium.
•
To introduce the concept of the free-body diagram.
•
To show how to solve rigid-body equilibrium problems in two
and three dimensions.
4. 1
IF,
CONDITIONS FOR RIGID-BODY
EQUILIBRIUM
In this section, we will develop both the necessary and sufficient conditions
for the equilibrium of the rigid body shown in Fig. 4-la. This body is
subjected to an external force and couple moment system that is the result
of the effects of gravitational, electrical, magnetic, or contact forces caused
by supports or adjacent bodies. The internal forces caused by interactions
between particles within the body are not shown in this figure, because
these forces occur in equal but opposite collinear pairs and hence will
cancel out, a consequence of Newton's third Jaw.
o.
f\
"11\
..,-M2
Fig. 4-1
157
158
CHAPTER
4
IF,
F4
"
F3
--
o.
- {)
..,..M2
Using the methods of the previous chapter, the force and couple moment
system acting on a body can be reduced to an equivalent resultant force and
resultant couple moment at any arbitrary point 0 on or off the body,
Fig. 4-lb. If these two resultants are both equal to zero, then the body is
said to be in equilibrium, which means it is at rest or will move with constant
velocity. Mathematically, the equilibrium of a body is expressed as
FR = 2 F
/
M1
EQUILIBRIUM OF A RIG I D BODY
= 0
\ F2
w
(MR)o
(a)
=
2M0
(4-1)
= 0
The first of these equations states that the sum of the forces acting on the
body is equal to zero. The second equation states that the sum of the
moments of all the forces in the system about point 0 , added to all the
couple moments, is equal to zero. These two equations are not only necessary
for equilibrium, they are also sufficient. To show this, consider summing
moments about some other point, such as point A in Fig. 4-lc. We require
(b)
Since r ~ 0, this equation is satisfied if Eqs. 4-1 are satisfied, namely
FR= 0 and (MR)o = 0.
When applying the equations of equilibrium, we will assume that the
body remains rigid. In reality, all bodies deform when subjected to loads;
however, most engineering materials such as steel and concrete are very
stiff and so their deformation is usually very small. Therefore, when
applying the equations of equilibrium, we can generally assume that the
body will remain rigid and not deform under the applied load without
introducing any significant error. This way the direction of the applied
forces and their moment arms with respect to a fixed reference remain
the same both before and after a load is applied.
r
A
(c)
Fig. 4-1 (cont.)
w
oooocc
2T
Fig. 4-2
EQUILIBRIUM IN TWO DIMENSIONS
In the first part of the chapter, we will consider the case where the force
system acting on a rigid body lies in or may be projected onto a single plane
and, furthermore, any couple moments acting on the body are directed
perpendicular to this plane. This type of force and couple system is often
referred to as a two-dimensional or coplanar force system. For example,
the airplane in Fig. 4-2 has a plane of symmetry through its center axis, and
so the loads acting on the airplane are symmetrical with respect to this
plane. Thus, each of the two wing tires will support the same load T, which
is represented on the side (two-dimensional) view of the plane as 2T .
4.2
4.2
15 9
FREE· BODY DIAGRAMS
FREE-BODY DIAGRAMS
Successful application of the equations of equilibrium, which will be
discussed in Sec. 4.3, requires a complete specification of all the known
and unknown exte rnal forces that act on the body. The best way to account
for these forces is to draw a f ree-body diagram of the body. This diagram
is a sketch of the o utlined shape of the body, which represents it as being
isolated or "free" from its surroundings, i.e., a "free body." On this sketch
it is necessary to show all the forces and couple moments that the supports
and the surroundings exert on the body, so that these effects can be
accounted for when the equations of equilibrium are applied. A tho rough
understanding of how to draw a free-body diagram is ofprimary impo rtance
for solving problems in both statics and mechanics of materials.
Support Reactions Before presenting a formal procedure as to
how to draw a free-body diagram, we will first consider the various types
of reactions that occur at supports and at points of contact between
bodies subjected to coplanar force systems. As a general rule,
•
A support pre vents the translation of a body by exerting a force o n
the body.
roller
F
•
(a)
A support prevents the rotation of a body by exerting a couple
moment on the body.
For example, let us consider three ways in which a horizontal member,
such as a beam, is suppo rted at its e nd. One method consists of a ro ller or
cylinder, Fig. 4-3a. Since this support only prevents the beam from
translating in the vertical direction, the roller will only exert a force on
the beam in this direction, Fig. 4-3b.
The beam can be supported in a more restrictive manner by using a
pin, Fig. 4-3c. The pin passes through a hole in the beam and two leaves
which are fixed to the ground. Here the pin can prevent translation of the
beam in any direction</>, Fig. 4-3d, and so the pin must exert a force F on
the beam in the opposite direction. For purposes of analysis, it is generally
easier to represent this resultant force F by its two rectangular
components f'.t and F,,, Fig. 4-3e. Once flx and F,, are known, then F and <f>
can be calculated.
The most restrictive way to support the beam would be to use a fixed
support as shown in Fig. 4-3[ This support will prevent both translation
and rotation of the beam. As a result, a force and couple moment must be
developed on the beam at its point of connection, Fig. 4-3g. Like the case
of the pin, the force is usually represented by its rectangular components
F:t and Fy.
Table 4-1 lists other common types of supports for bodies subjected to
coplanar force syste ms. (In all cases the angle 8 is assumed to be known.)
Carefully study each of the symbols used to represent these supports and
the types of reactions the y exert on their contacting members.
(b)
ll member
.... pin
• UJ. leaves
.'&
1
_L_
pin
(c)
or
F,
(d)
(e)
(g)
(f)
Fig. 4-3
160
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
TABLE 4-1 Supports for Rigid Bodies Subjected to Two-Dimensional Force Systems
Types of Connection
Reaction
Number of Unknowns
(1)
~
~
One unknown. The reaction is a tension force which acts
away from the member in the direction of the cable.
cable
(2)
or
One unknown. The reaction is a force which acts along
the axis of the link.
weightless link
(3)
;/
roller
(4)
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
F
~
8("Z
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
F
rocker
(5~
smooth contacting
surface
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
F
(6)
orF~
One unknown. The reaction is a force which acts
perpendicular to the slot.
roller or pin in
confined smooth slot
(7)
One unknown. The reaction is a force which acts
perpendicular to the rod.
member pin connected
to collar on smooth rod
continued
4.2
FREE·BODY DIAGRAMS
161
TABLE 4-1 Continued
.t
Types of Connection
(8)
'
Reaction
or
Number of Unknowns
Two unknowns. The reactions are two components of
force, or the magnitude and direction 4> of the resultant
force. Note that 4> and 8 are not necessarily equal [usually
not, unless the rod shown is a link as in (2)].
smooth pin or hinge
(9)
Two unknowns. The reactions arc the couple moment and
the force which acts perpendicular to the rod.
member fixed connected
to collar on smooth rod
(Hl)
Three unknowns. The reactions arc the couple moment
and the two force components, or the couple moment and
the magnitude and direction <J> of the resultant force.
fixed support
Typical examples of actual supports are shown in the following sequence of photos. The numbers refer to the
connection types in Table 4-1.
This concrete girder
rests on the ledge that
is assumed to act as
a smooth contacting
surface. (5)
The cable exerts a force on the bracket
in the direction of the cable. (1)
Typical pin support for a beam. (8)
The rocker support for this
bridge girder allows horizontal
movement so the bridge is free
to expand and contract due to
a change in temperature. ( 4)
The Ooor beams of this
building are welded
together and thus form
fixed connect ions. (I 0)
162
CHAPTER
4
EQUILIBRIUM OF A RIG I D BODY
Springs. If a linear elastic spring as in Fig. 4-4 is used to support a body,
the length of the spring will change in direct proportion to the force acting
on it. A characteristic that defines the "elasticity" of a spring is the spring
constant or stiffness k. Specifically, the magnitude of force developed by a
linear elastic spring which has a stiffness k, and is deformed (elongated or
compressed) a distances measured from its unloaded position, is
(4-2)
Note that s is determined from the difference in the spring's deformed
length land its undeformed length 10 , i.e.,s = l - 10 .
F
Weight and the Center of Gravity. When a body is within a
Fig. 4-4
gravitational field, then each of its particles has a specified weight. It was
shown in Sec. 3.8 that such a system of forces can be reduced to a single
resultant force acting through a specified point. We refer to this force
resultant as the weight W of the body and to the location of its point of
application as the center ofgravity. The methods used for its determination
will be developed in Chapter 6. In the examples and problems that follow,
if the weight of the body is important for the analysis, this force will be
reported in the problem statement.
Internal Forces. As stated in Sec. 4.1, the internal forces that act
w
(b)
(a)
Fig. 4-5
between adjacent particles in a body always occur in collinear pairs such
that they have the same magnitude and act in opposite directions (Newton's
third Jaw). Since these forces cancel each other, they will not create an
external effect on the body. It is for this reason that the internal forces should
not be included on the free-body diagram if the entire body is to be
considered. For example, the engine shown in Fig. 4-Sa has a free-body
diagram shown in Fig. 4-Sb. The internal forces between all its connected
parts, such as the screws and bolts, will cancel out. Only the external forces
T1 and T2 exerted by the chains and the engine weight W are shown on the
free-body diagram.
Idealized Models. When an engineer performs a force analysis of
any object, he or she must consider a corresponding analytical or
idealized model that gives results that approximate as closely as possible
the actual situation. To do this, careful choices have to be made so that
selection of the type of supports, the material behavior, and the object's
dimensions can be justified. This way one can feel confident that any
design or analysis will yield results which can be trusted. In complex
cases this process may require developing several different models of the
object that must be analyzed. However, in any case, this selection process
requires both skill and experience.
4.2
The following two cases illustrate what is required to develop a proper
model. In Fig. 4-6a, the steel beam is to be used to support the three roof
joists of a building. For a force analysis it is reasonable to assume the
material (steel) is rigid since only very small deflections will occur when
the beam is loaded. A bolted connection at A will allow for any slight
rotation that occurs here when the load is applied, and so a pin can be
considered for this support. At B a roller can be considered since this
support offers no resistance to horizontal movement. A building code is
used to specify the roof loading so that the joist loads F can be calculated.
These forces are intented to be larger than any actual loading on the
beam since they account for extreme loading cases and for any dynamic
or vibrational effects. Finally, the weight of the beam is generally
neglected when it is small compared to the load the beam supports.
The idealized model of the beam is therefore shown with average
dimensions a, b, c, and din Fig. 4-6b.
As a second case, consider the lift boom in Fig. 4-7a. By inspection, it is
supported by a pin at A and by the hydraulic cylinder BC, which can be
approximated as a weightless link. The material can be assumed rigid,
and with its density known, the weight of the boom and the location of its
center of gravity Gare determined. When a design loading P is specified,
the idealized model shown in Fig. 4-7b can be used for a force analysis.
Average dimensions (not shown) are used to specify the location of the
loads and the supports.
Idealized models of specific objects will be given in some of the
examples throughout the text. In all cases, it should be realized that each
represents the reduction of a practical situation using simplifying
assumptions Like the ones illustrated here.
p
(b)
(a)
Fig. 4-7
FREE-BODY D IAGRAMS
(a)
(b)
Fig. 4-6
163
164
CHAPTER
4
EQUILIBRIUM OF A RIG I D BODY
IMPORTANT POINTS
• No equilibrium problem should be solved without first drawing the free-body diagram, so as to account
for all the forces and couple moments that act on the body.
• If a support prevents translation of a body, then the support exerts a force on the body.
• If a support prevents rotation of a body, then the support exerts a couple moment on the body.
• The force Fin an elastic spring is related to the extension or compression of the spring using F = ks,
where k is the spring's stiffness.
• The weight of a body is an external force, and its effect is represented by a single resultant force acting
through the body's center of gravity G.
• Internal forces are never shown on the free-body diagram since they occur in equal but opposite collinear
pairs and therefore cancel out.
• Couple moments can be placed anywhere on the free-body diagram since they are free vectors. Forces
can act at any point along their lines of action since they are sliding vectors.
PROCEDURE FOR ANALYSIS
To construct a free-body diagram for a rigid body or any group of bodies considered as a single system, the
following steps should be performed:
Draw Outlined Shape.
Imagine the body to be isolated or cut "free" from its constraints and connections and draw (sketch) its
outlined shape.
Show All Forces and Couple Moments.
Identify all the known and unknown external forces and couple moments that act on the body. Those generally
encountered are due to (1) applied loadings, (2) reactions occurring at the supports or at points of contact
with other bodies, and (3) the weight of the body. To account for all these effects, it may help to trace over the
boundary, carefully noting each force or couple moment acting on it.
Identify Each Loading and Give Dimensions.
The forces and couple moments that are known should be labeled with their proper magnitudes and directions.
Letters are used to represent the magnitudes and direction angles of forces and couple moments that are
unknown. Finally, indicate the dimensions of the body necessary for calculating the moments of forces.
4.2
EXAMPLE
FREE-BODY DIAGRAMS
1 65
4 .1
The sphere in Fig. 4-& has a mass of 6 kg and is supported as shown.
Draw a free-body diagram of the sphere, the cord CE, and the knot at C.
Fe£ (Force of cord CE acting OD sphere)
B
k
D
58.9 N (Weight or gravity acting OD sphere)
(a)
SOLUTION
(b)
F EC (Force of knot acti ng o n cord CE)
Sphere. By inspection, there are only two forces acting on the sphere,
namely, its weight, 6 kg (9.81 m/s2) = 58.9 N, and the force of cord CE.
The free-body diagram is shown in Fig. 4-8b.
When the cord CE is isolated from its surroundings, its freebody diagram shows only two forces acting on it, namely, the force of the
sphere and the force of the knot, Fig. 4-8c. Notice that Fe£ shown here is
equal but opposite to that shown in Fig. 4-8b, a consequence of Newton's
third law of action-reaction.Also, Fe£ and F£c pull on the cord and keep Fe£ (Force of sphere acting on cord CE)
it in tension so that it doesn't collapse. For equilibrium, Fe£= FEC·
Cord CE.
(c)
Knot. The knot at C is subjected to three forces, Fig. 4-8d. They are
caused by the cords CBA and CE and the spring CD. As required, the
free-body diagram shows all these forces labeled with their magnitudes
and directions. It is important to recognize that the weight of the sphere
does not directly act on the knot. Instead, the cord CE subjects the knot
to this force.
FCBA (Force of cord CBA acting on knot)
c
,..__ ___ F co (Force of spring acting on knot)
Fe£ (Force of cord CE acting on knot)
(d)
Fig. 4-8
166
I
CHAPTER
EXAMPLE
4
EQUILIBRIUM OF A RIG I D BODY
4.2
Draw the free-body diagram of the foot lever shown in Fig. 4-9a. The
operator applies a vertical force to the pedal so that the spring is stretched
1.5 in. and the force in the short link at B is 20 lb.
I
1.5 in.
I
1 in.
(b)
t
1.yn.
1 in.
(a)
Fig. 4-9
SOLUTION
By inspection of the photo, the lever is loosely bolted to the frame at A.
The rod at Bis pinned at its ends and acts as a "short link." After making
the proper measurements, the idealized model of the lever is shown in
Fig. 4-9b. From this, the free-body diagram is shown in Fig. 4-9c. The pin
support at A exerts force components Ax and Ayon the lever. The link at
B exerts a force of 20 lb, acting in the direction of the link. In addition the
spring also exerts a horizontal force on the lever. If the stiffness is
measured and found to be k = 20 lb/in., then since the stretch
s = 1.5 in., using Eq. 4-2, F, = ks = 20 lb/in. (1.5 in.) = 30 lb. Finally,
the operator's shoe applies a vertical force of F on the pedal. The
dimensions of the lever are also shown on the free-body diagram, since
this information will be useful when calculating the moments of the
forces. As usual, the senses of the unknown forces at A have been
assumed. The correct senses will become apparent after solving the
equilibrium equations.
4.2
EXAMPLE
16 7
FREE· BODY DIAGRAMS
4 .3
Two smooth pipes, each having a mass of 300 kg, are supported by the
forked tines of the tractor in Fig. 4-lOa. Draw the free-body diagrams for
each pipe and both pipes together.
Effect of B acting on A
R
300
Effect of sloped
blade acting on A
(a)
(b)
2943 N
Effect of gravi ty
(weight) acting on A
(c)
F
Effect of sloped
fork acting on A
SOLUTION
The idealized model from which we must draw the free-body diagrams is
shown in Fig. 4-IOb. H ere the pipes are identified, the dimensions have
been added, and the physical situation is reduced to its simplest form.
The free-body diagram of pipe A is shown in Fig. 4-lOc. Its weight is
W = 300(9.81) N = 2943 N. Assuming all contacting surfaces are
smooth, the reactive forces T ,F, R act in a directionnonna/ to the tangent
at their surfaces of contact.
The free-body diagram of pipe B is shown in Fig. 4-lOd. Can you
identify each of the three forces acting on this pipe? Note that R
representing the force of A on B, Fig. 4-lOd, is equal and opposite to R
representing the force of Bon A, Fig. 4-lOc.
The free-body diagram of both pipes combined ("system") is shown in
Fig. 4-IOe. H e re the contact force R , which acts between A and B , is
considered an internal force and hence is not shown on the free-body
diag ram. That is, it represents a pair of equal but opposite collinear forces
whic h cancel each othe r.
R
p
(d)
T
F
(e)
Fig. 4-10
168
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
CONCEPTUAL PROBLEMS
C4-L Draw the free-body diagram of the uniform trash
bucket which has a significant weight. It is pinned at A and
rests aganist the smooth horizontal member at B. Show your
result in side view. Label any necessary dimensions.
C4-3. Draw the free-body diagram of the wing on the
passenger plane. The weights of the engine and wing are
significant. The tires at B are smooth.
Prob. C4-3
Prob. C4-1
C4-2. Draw the free-body diagram of the outrigger ABC
used to support a backhoe. The top pin B is connected to the
hydraulic cylinder, which can be considered to be a short link
(two-force member), the bearing shoe at A is smooth, and
the outtrigger is pinned to the frame at C.
Prob. C4-2
C4-4. Draw the free-body diagram of the wheel and
member ABC used as part of the landing gear on a jet plane.
The hydraulic cylinder AD acts as a two-force member, and
there is a pin connection at B.
Prob. C4-4
4.3
4. 3
EQUATIONS OF EOUIUBRIUM
EQUATIONS OF EQUILIBRIUM
In Sec. 4.1 we developed the two equations which are both necessary and
sufficient for the equilibrium of a rigid body, namely, I F = 0 and
I M 0 = 0. When the body is subjected to a system of forces, which all lie
in the x-y plane, then the forces can be resolved into their x and y
components. The conditions for equilibrium in two dimensions then
become
IF.x
IF.)'
IM0
= O
= O
=0
(4-3)
Here !f"x and I Fy represent, respectively, the algebraic sums of the x and
y components of all the forces acting on the body, and IM0 represents
the algebraic sum of the couple moments and the moments of all th,e force
components about the z axis that pass through the arbitrary point 0.
(a)
Alternative Sets of Equilibrium Equations. Although Eqs. 4-3
are most often used for solving coplanar equilibrium problems, two
alternative sets of three independent equilibrium equations may also be
used. One such set is
IF.x = O
IMA= 0
IMB= 0
8 lo<'-- - - - - --' c
(4-4)
When using these equations, it is required that the line passing through
points A and B not be parallel to they axis. To show that these equations
provide the conditions for equilibrium, consider the free-body diagram
of the plate in Fig. 4-lla. Using the methods of Sec. 3.7, all the forces on
the free-body diagram are first replaced by an eq_uivalent resultant force
FR= I F, and a resultant couple moment ( MR)A = I MA, Fig. 4-llb.
lf IMA = 0 is satisfied, then ( MR) A = 0. lf If"x = 0 is satisfied, then
FR must have no component along the x axis, and therefore FR must be
parallel to they axis, Fig. 4-llc. Fmally, if !Ma = 0, where B does not lie
on the line of action of FR, then FR = 0 and therefore the body in
Fig. 4-1 la must be in equilibrium.
(b)
r.F=;=====---~ A
8 lo<'-- - - - - --' c
(c)
Fig. 4-11
169
170
CHAPTER
4
EQUILIBRIUM OF A RIG I D BODY
A second alternative set of equilibrium equations is
2MA = 0
'I.Ms = 0
'I.Mc = 0
(a)
(4-5)
Here it is necessary that points A, B, and C do not lie on the same line. To
show that these equations, when satisfied, ensure equilibrium, consider
again the free-body diagram in Fig. 4- llb. If 2MA = 0 is to be satisfied,
then ( MR ) A = 0. If 2Mc = 0 is satisfied, then the line of action of FR
passes through point C, Fig. 4-llc. Finally, if 2M8 = 0 is satisfied, then
FR = 0, and so the plate in Fig. 4-lla must be in equilibrium.
PROCEDURE FOR ANAL YS/S
Coplanar force equilibrium problems can be solved using the
following procedure.
n""---- - - ------ c
(b)
Free-Body Diagram.
• Establish the x, y coordinate axes in any suitable orientation.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on the body.
,...~---------:>I A
• Label all the loadings and specify their directions relative to the x
or y axis. The sense of a force or couple moment having an
unknown magnitude but known line of action can be assumed.
• Indicate the dimensions of the body necessary for calculating the
moments of forces.
n""---- - - ------ c
(c)
Fig. 4-11 (Repeated)
Equations of Equilibrium.
• Apply the moment equation of equilibrium, 2M0 = 0, about
a point 0 that lies at the intersection of the lines of action of
two unknown forces. In this way, the moments of these
unknowns are zero about 0 , and a direct solution for the third
unknown can be determined.
• When applying the force equilibrium equations, 24 = 0 and
2F,; = 0, orient the x and y axes along lines that will provide the
sunplest resolution of the forces into their x and y components.
• If the solution of the equilibrium equations yields a negative
scalar for a force or couple moment magnitude, this indicates
that the sense is opposite to that which was assumed on the
free-body diagram.
4.3
EXAMPLE
EQUATIONS OF EOUIUBRIUM
1 71
4.~
Determine the tension in cables BA and BC necessary to support the 60-kg
cylinder in Fig. 4-12.a.
T 80 = 60 (9.81) N
B
D
60 (9.81) N
(a)
(b)
SOLUTION
Due to equilibrium, the weight of the cylinder
causes the tension in cable BD to be T80 = 60(9.81) N, Fig. 4-12b. The
forces in cables BA and BC can be determined by investigating the
equilibrium of ring B. Its free-body diagram is shown in Fig. 4-12c. The
magnitudes of TA and Tc are unknown, but their directions are known.
Free-Body Diagram.
y
Equations of Equilibrium. Applying the equations of equilibrium
along the x and y axes, we have
~I~= 0;
+f l£,.
= 0;
Tccos45° - (~) TA = 0
Tc sin 45° +
(D TA
- 60(9.81) N = 0
(1)
(2)
Equation (1) can be written as TA = 0.8839Tc- Substituting this into
Eq. (2) yields
Tc sin 45° + (~}(0.8839Tc) - 60(9.81)N = 0
So that
Tc = 475.66 N = 476 N
Substituting this result into either Eq. (1) or Eq. (2), we get
TA
= 420 N
T 8 v = 60 (9.81) N
Ans.
Ans.
NOTE: The accuracy of these results, of course, depends on the accuracy
of the data, i.e., measurements of geometry and loads. For most
engineering work involving a problem such as this, the data as measured
to three significant figures would be sufficient.
(c)
Fig. 4-12
172
CHAPTER
-EXAMPLE
4
4.5
EQUILIBRIUM OF A RIG ID BODY
Determine the horizontal and vertical components of reaction on the beam
caused by the pin at Band the rocker at A as shown in Fig. 4-13a. Neglect
the weight of the beam.
y
> 0.2m
600N
t
Aiji ~ . fI
4
I
600 sin 45° N
200N
200N
?? ' n
i:mJ1-3m-1~zT
-2m~--3m--D-•-2mlOON
lOON
(b)
(a)
Fig. 4-13
SOLUTION
Identify each of the forces shown on the freebody diagram in Fig. 4-13b. Here the 600-N force is represented by its
x and y components.
Free-Body Diagram.
Equations of Equilibrium.
~ 2-Fr = O;
Summing forces in the x direction yields
600cos45°N - Bx = 0
Bx = 424 N
Ans.
A direct solution for Ay can be obtained by applying the moment
equation about point B.
~+2.Ms = O;
100 N (2 m) + (600 sin 45° N)(5 m)
- (600 cos 45° N)(0.2 m) - Ay(7 m)
Ay
1319 N
A (beam on pin)
t
319N
(pin on rocker)
(c)
319N
(floor on rocker)
O
319 N
Ans.
Summing forces in they direction, using this result, gives
+f 2£,, =
0;
319N - 600sin45°N - lOON - 200N + By
By
t 319N
(rocker on pin)
=
=
=
405 N
=
0
Ans
NOTE: The support forces in Fig. 4- 13b are caused by the pins that act on
the beam. The opposite forces act on the pins. For example, Fig. 4-13c
shows the equilibrium of the pin at A and the rocker.
4.3
EXAMPLE
17 3
EQUATIONS OF EOUIUBRIUM
4 .6
The cord shown in Fig. 4-14a supports a force of 100 lb and wraps over the
frictionless pulley. De termine the tension in the cord at C and the horizontal
and vertical components of reaction at pin A .
100 lb
(a)
Fig. 4-14
SOLUTION
p
Free-Body Diagrams. The free-body diagrams of the cord and pulley
are shown in Fig. 4-14b. Note that the principle of action, equal but
opposite reactio n must be carefully observed when drawing each of
these diagrams: the cord exe rts an unknown load distribution p on the
pulley at the contact surface, whereas the pulley exerts an equal but
opposite effect on the cord. For the solution, however, it is simpler to
combine the free-body diagrams of the pulley and this portion of the
cord, so that the distributed load becomes internal to this ..system" and is
therefore e liminated fro m the analysis, Fig. 4-14c.
100
lb
T
(b)
0.5 fl
Summing moments about point A to
e liminate Ax a nd Ay, Fig. 4-14c, we have
Equations of Equilibrium.
100 lb (0.5 ft) - T(0.5 ft)
0
=
T = 100 lb
Ans.
Using this result,
..:!; ~F,r
= O;
-Ax + 100 sin 30° lb
=
IOOlb
0
(c)
Ax= 50.0 lb
+ f ~F,. = O;
Ay - 100 lb - 100 cos 30° lb
Ay = 187 lb
T
Ans.
=
0
Ans.
NOTE: It is seen that the tension in the cord remains constant as the cord
passes over the pulley. (This of course is true for any angle 8 at which the
cord is directed a nd for any radius r of the pulley.)
174
CHAPTER
-EXAMPLE
4
4.7
EQUILIBRIUM OF A RIG I D BODY
The member shown in Fig. 4-15a is pin connected at A and rests against a
smooth support at B. Determine the horizontal and vertical components of
reaction at the pin A.
B
-L------>{
~
v-----~o)
90N · m
(a)
(b)
Fig. 4-15
SOLUTION
Free-Body Diagram. As shown on the free-body diagram, Fig. 4-15b,
the reaction Na must be perpendicular to the member at B. Also,
horizontal and vertical components of reaction are represented at A.
Equations of Equilibrium.
direct solution for Na,
C +lMA
=
O;
Summing moments about A, we obtain a
-90N·m - 60N(lm) + Na(0.75m)
=
0
Na = 200N
Using this result,
~IF.x =
+ jlF.y
=
O·,
O·,
Ax - 200 sin 30° N = 0
Ax = lOON
Ay - 200 cos 30° N - 60 N = 0
A>' = 233 N
Ans.
Ans.
4.4
Two- AND THREE-FORCE MEMBERS
17 5
4.4 TWO- AND THREE-FORCE
MEMBERS
The solutions to some equilibrium problems can be simplified by
recognizing members that are subjected to only two or three forces.
Two-Force Members. As the name implies, a two-force member has
forces applied at only two points on the member. An example of a twoforce member is shown in Fig. 4-16a. To satisfy force equilibrium, FA and
Fa must be equal in magnitude, FA = Fa = F, but opposite in direction
(:~F = 0), Fig. 4-16b. Furthermore, moment equilibrium requires that FA
and Fa share the same line of action, which can only happen if they are
directed along the line joining points A and B (IMA = 0 or lM8 = 0),
Fig. 4- 16c. Therefore, for any two-force member to be in equilibrium, the
two forces acting on the member must have the same magnitude, act in
opposite directions, and have the same line of action, directed along the Line
joining the two points where these forces act.
F8 =F
(a)
(b)
(c)
Two-force member
Fig. 4-16
The hydraulic cylinder AB is a typical
example of a two-force member since it
is pin connected at its ends and, provided
its weight is neglected, only the resultant
pin forces act on this member.
The boom-and-bucket on this lift is a
three-force member, provided its weight is
neglected. Here the lines of action of the
weight of the worker, W, and the force of
the two-force member (hydraulic cylinder)
at B, F8 , intersect at 0. For moment
equilibrium, the resultant force at the pin
A , FA, must also be directed towards 0.
Three-Force Members. If a member is subjected to only three forces,
it is called a three-force member. Moment equilibrium can be satisfied only
if the three forces form a concurrent or parallel force system. To illustrate,
consider the member in Fig. 4-17a subjected to the three forces FI> !F2, and
F3 . If the lines of action of F1 and F2 intersect at point 0 , then the line of
action of F3 must also pass through point 0 so that the forces satisfy
l M 0 = 0. As a special case, if the three forces are all parallel, Fig. 4-17b,
the location of the point of intersection, 0, will approach infinity.
0
(a)
(b)
Three-force member
Fig. 4-17
The link used for this railroad car brake
is a three-force member. Since the force
F8 in the tie rod at B and Fe from the
link at Care parallel, then for equilibrium
the resultant force FA at the pin A must
also be parallel with these two forces.
176
4
CHAPTER
-EXAMPLE
EQUILIBRIUM OF A RIG I D BODY
4.8
The lever ABC is pin supported at A and connected to a short link BD as
shown in Fig. 4-18a. If the weight of the members is negligible, determine
the force of the pin on the lever at A.
SOLUTION
0.5 m
Free-Body Diagrams. As shown in Fig. 4-18b, the short link BD is
a two-force member, so the resultant forces from the pins D and B must
~
0.201
be equal, opposite, and collinear. Although the magnitude of the force is
unknown, the line of action is known since it passes through Band D.
Lever ABC is a three-force member, and therefore, in order to
satisfy moment equilibrium, the three nonparallel forces acting on it
must be concurrent at 0, Fig. 4- 18c. Note that the force Fon the lever at
B is equal but opposite to the force F acting at B on the link. Why? The
distance CO must be 0.5 m since the line of action of F is known.
r---i ~
I
0.2m
_L
•D
.•
r·
0.1 m
(a)
Equations of Equilibrium. By requiring the force system to be
concurrent at 0 , since IM0 = 0, the angle 8 which defines the line of
F
action of FA can be determined from trigonometry,
[
F
45° ' \
I
D
8 = tan- 1(
I
I
I
/
I
(b)
I
Using the x, y axes and applying the force equilibrium equations,
I
I
I
I
I
' c
07
· ) = 60.3°
0.4
o.sm - J
400N
I
I
I
I
\
1
0
~IF.x
=
O·,
FA cos 60.3° - F cos 45° + 400 N
+ jIF.y
=
O·,
FA sin 60.3° - F sin 45°
=
=
0
O
Solving, we get
'L
FA = 1.07 kN
F
=
Ans.
1.32kN
1 45•\
0.2m
I
NOTE: We can also solve this problem by representing the force at A by
F
its two components Ax and Ay and applying IMA = 0 to get F, then
IF., = 0, IF,. = 0 to get Ax and A y. Once Ax and Ay are determined, we
can get FA and 8.
(c)
Fig. 4-18
4.4
17 7
Two- AND THREE-FORCE M EMBERS
PRELIMINARY PROBLEM
P4-l .
Draw the free-body diagram of each object.
SOON
i---4m---1
SOON
b---------' B
_..~..._ 3 m--1-2m ~
(d)
(a)
- :;:ttWJrN.
.A~•
m
l -2m
3m
B
l
A
,-----.,...----~B
2 m-~2 m--I
(e)
(b)
400N/m
,~.
•
-3m~-3m-J
A
(c)
-.....,2-m-B-1l~C
i
l -
(f)
Prob. P4-l
178
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
FUNDAMENTAL PROBLEMS
All problem solutions must include an FBD.
F4-1. Determine the horizontal and vertical components
of reaction at the supports. Neglect the thickness of the
beam.
F4-4. Determine the components of reaction at the fixed
support A. Neglect the thickness of the beam.
200 N
200 N
200 N
500Jb
~
6001b·ft
'!f\.
() B ;i)
1-sft - l -sft -~s ft ~
Prob.F4-1
F4-2. Determine the horizontal and vertical components
of reaction at the pin A and the reaction on the beam at C.
11--l.Sm
1 - -1.Sm
-l o
4kN!
B
A
Prob.F4-4
F4-S. The 25-kg bar has a center of mass at G. If it is
supported by a smooth peg at C, a roller at A , and cord AB,
determine the reactions at these supports.
IA
1.5 m
B
Prob.F4-2
F4-3. The truss is supported by a pin at A and a roller
at B. Determine the support reactions.
Prob. F4-S
F4-6. Determine the reactions at the smooth contact
points A , B, and Con the bar.
Prob.F4-3
Prob.F4-6
4.4
179
Two- AND THREE-FORCE MEMBERS
PROBLEMS
All problem solutions must include an FBD.
4-L Determine the components of the support reactions
at the fixed support A on the cantilevered beam.
*4-4.
Determine the reactions at the supports.
900N/m
r--,.._
--.,.._
-
--.,.._
600N/m
6kN
A
~ !1
•
---3m ~1--- 3n1---1
Prob. 4-4
Prob. 4-1
4-5. Determine the reactions at the supports.
4-2. Determine the reactions at the supports.
400 N/m
3 n1
1 - - -3 m - -·1- - -3 m - -- 1
-
1 m - - i-- - - - 3 m - - - - 1
Prob. 4-2
Prob. 4-5
4-3. Determine the horizontal and vertical components
of reaction of the pin A and the reaction of the rocker B on
the beam.
4-6. Determine the reactions at the supports.
Sk~
N~>(JJ.====;;;;;:;~:;-~~~-,1
4kN
A
2m
~~Y=~··=~:'sl
o
6kN
1 - - - - 6 m - - - -- 1
Prob. 4-3
- - 2m
8 kN
--~ 2m ~-- 2m - -·
Prob. 4-6
180
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
4-7. Determine the magnitude of force at the pin A and in
the cable BC needed to support the 500-lb load. Neglect the
weight of the boom AB.
4-10. The smooth pipe rests against the opening at the
points of contact A , B, and C. Determine the reactions at
these points needed to support the force of 300 N. Neglect
the pipe's thickness.
B
A
Prob. 4-7
*4-8. The dimensions of a jib crane are given in the figure.
If the crane has a mass of 800 kg and a center of mass at G,
and the maximum rated force at its end is F = 15 kN,
determine the reactions at its bearings. The bearing at A is a
journal bearing and supports only a horizontal force,
whereas the bearing at B is a thrust bearing that supports
both horizontal and vertical components.
4-9. The dimensions of a jib crane are given in the figure. The
crane has a mass of 800 kg and a center of mass at G. The
bearing at A is a journal bearing and can support a horizontal
force, whereas the bearing at B is a thrust bearing that supports
both horizontal and vertical components. Determine the
maximum load F that can be suspended from the end of the
crane if the bearings at A and B can sustain a maximum
resultant load of24 kN and 34 kN, respectively.
c
/I
O5 m-
~· .
30•(._
-+-- - 0.5 m
Jo.J6 m
_j
B
--1--l
0.15m
300N
Prob. 4-10
4-11. The beam is horizontal and the springs are
unstretched when there is no load on the beam. Determine
the angle of tilt of the beam when the load is applied.
3m
A
0.75m
B
2m
l
600N/m
B
--3m
Probs. 4-819
k 8 = 1.5 kN/m
--1-- --1
3m
Prob. 4-11
4.4
*4-U. The 10.kg uniform rod is pinned at end A. If it is
subjected to a couple moment of 50 N · m, determine the
smallest angle e for equilibrium. The spring is unstretched
when e = 0°, and has a stiffness of k = 60 N/ m.
~ k~~Nlm
0
181
Two- AND THREE-FORCE MEMBERS
4-15. Determine the reactions at the pin A and the tension
in cord BC. Set F= 40 kN. Neglect the thickness of the beam.
*4-16. If rope BC will fail when the tension becomes 50 kN,
determine the greatest vertical load F that can be applied to
the beam at B. What is the magnitude of the reaction at A for
this loading? Neglect the thickness of the beam.
B
2m
~ o.sm -
F
26kN
t::Bf
c
50N · m
Prob. 4-12
4-13. The man uses the hand truck to move material up
the step. If the truck and its contents have a mass of 50 kg
with center of gravity at G, determine the normal reaction
on both wheels and the magnitude and direction of the
minimum force required at the grip B needed to lift the load.
A
~ 2m -1--- 4m --~1
Probs. 4-15n6
4-17. The rigid metal strip of negligible weight is used as
part of an electromagnetic switch. If the stiffness of the
springs at A and B is k = 5 N/ m and the strip is originally
horizontal when the springs are unstretched, determine the
smallest force F needed to close the contact gap at C.
Prob. 4-13
4-14. Three uniform books, each having a weight W and
length a, are stacked as shown. Determine the maximum
distanced that the top book can extend out from the bottom
one so the stack does not topple over.
50 mm - - - -50 mm - - - 1
!F
?pi
- - - -a - - - - + -d Prob. 4-14
Prob. 4-17
c:::J.f...
_lomm
=
182
C H APT ER
4
EQU ILI BR I UM OF A RI GID BODY
4-18. The rigid metal strip of negligible weight is used as
part of an electromagnetic switch. Determine the maximum
stiffness k of the springs at A and B so that the contact at C
closes when the vertical force developed there is F = 0.5 N.
Originally the strip is horizontal as shown.
SOmm- J
*4-20. The uniform beam has a weight Wand length I and
is supported by a pin at A and a cable BC. Determine the
horizontal and vertical components of reaction at A and the
tension in the cable necessary to hold the beam in the
position shown.
SOmm- J
t
!F
A•
c::::J c
/!&
r omm
=-
Prob. 4-20
Prob. 4-18
4-19. The cantilever footing is used to support a wall near
its edge A so that it causes a uniform soil pressure under the
footing. Determine the uniform distribution loads, wA and
w8 , measured in lb/ft at pads A and B, necessary to support
the wall forces of 8000 lb and 20 000 lb.
4-2L A boy stands out at the end of the diving board, which
is supported by two springs A and B, each having a stiffness
of k = 15 kN / m. In the position shown the board is horizontal.
If the boy has a mass of 40 kg, determine the angle of tilt
which the board makes with the horizontal after he jumps off
Neglect the weight of the board and assume it is rigid.
20 000 lb
1!
8000 lb
- 025ft
-
,....
>---
I
~
1.5 f
I ·1
I
AW11
I
I
I
I
I
t t t t t tB
l-2 I
wAft
i---- - - 3 fl) - - - - -
,...- - - -8 ft - - - - - - i -3 wn
ft-
Prob. 4-19
Prob. 4-21
4.4
4-22. The beam is subjected to the two concentrated loads.
Assuming that the foundation exerts a linearly varying load
distribution on its bottom, determine the load intensities w 1
and w2 for equilibrium in terms of the parameters shown.
P
Two- AND THREE-FORCE M EMBERS
183
*4-24. Determine the distance d for placement of the load P
for equilibrium of the smooth bar when it is held in the position
8 as shown. Neglect the weight of the bar.
2P
r-+-i-+-i-~ --j
~
,,
'
l II
iv,
-
'
I•
-aProb. 4-24
Prob. 4-22
4-23. The rod supports a weight of200 lb and is pinned at its
end A. U it is subjected to a couple moment of 100 lb· ft,
determine the angle 8 for equilibrium. The spring has an
unstretched length of2 ft and a stiffness of k = 50 lb/ft.
4-25. U d = 1 m. and 8 = 30". determine the normal
reaction at the smooth supports and the required distance a
for the placement of the roller if P = 600 N. Neglect the
weight of the bar.
2 ft
k = 50 lb/ft
-aProb. 4-23
Prob. 4-25
184
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
CONCEPTUAL PROBLEMS
C4-5. The tie rod is used to support this overhang at the
entrance of a building. If it is pin connected to the building wall
at A and to the center of the overhang B, determine if the force
in the rod will increase, decrease, or remain the same if (a) the
support at A is moved to a lower position D,and (b) the support
at Bis moved to the outer position C. Explain your answer with
an equilibrium analysis, using dimensions and loads. Assume
the overhang is pin supported from the building wall.
C4-7. Like all aircraft, this jet plane rests on three wheels.
Why not use an additional wheel at the tail for better
support? (Can you think of any other reason for not
including this wheel?) If there was a fourth tail wheel, draw
a free-body diagram of the plane from a side (2 D) view, and
show why one would not be able to determine all the wheel
reactions using the equations of equilibrium.
Prob. C4-7
Prob. C4-5
C4-6. The man attempts to pull the four wheeler up the
incline and onto the trailer. From the position shown, is it
more effective to pull on the rope at A , or would it be better
to pull on the rope at B? Draw a free-body diagram for each
case, and do an equilibrium analysis to explain your answer.
Use appropriate numerical values to do your calculations.
Prob. C4-6
C4-8. Where is the best place to arrange most of the logs
in the wheelbarrow so that it minimizes the amount of force
on the backbone of the person transporting the load? Do an
equilibrium analysis to explain your answer.
Prob. C4-8
4.5
EQUILIBRIUM IN THREE DIMENSIONS
4. 5
FREE-BODY DIAGRAMS
The first step in solving three-dimensional equilibrium problems, as in
the case of two dimensions, is to draw a free-body diagram. Before we
can do this, however, it is first necessary to discuss the types of reactions
that can occur at the supports.
Support
React~ons. The reactive forces and couple moments acting
at various types of supports and connections, when the members are
viewed in three dimensions, are listed in Table 4-2. It is important to
recognize the symbols used to represent each of these supports and to
understand clearly how the forces and couple moments are developed.
As in the two-dimensional case:
•
A support prevents the translation of a body by exerting a force on
the body.
•
A support prevents the rotation of a body by exerting a couple moment
on the body.
For example, in Table 4-2, item (4), the ball-and-socket joint prevents
any translation of the connecting member; therefore, a force must act on
the member at the point of connection. This force has three components
having unknown magnitudes, f"x, £,,, F, . Provided these components are
known, one can obtain the magnitude of force, F = VF; + F~ + F~,
and the force's orientation defined by its coordinate direction angles a,
{3, -y, Eqs. 2-5. * Since the connecting member is allowed to rotate freely
about any axis, no couple moment is resisted by a ball-and-socket joint.
Notice that the single bearing supports in items (5) and (7), the single
pin (8), and the single hinge (9) are shown to resist both force and couplemoment components. 1~ however, these supports are used with other
bearings, pins, or hinges to hold a rigid body in equilibrium and these
supports are properly aligned when connected to the body, then the force
reactions at these supports alone are adequate for supporting the body.
In other words, the couple moments will not develop since the body is
prevented from rotating by the other supports. The reason for this should
become clear after studying the examples which follow.
*The three unknowns may also be represented as an unknown force magnitude F and
two unknown coordinate direction angles. The third direction angle is obtained using the
identity cos2 a + coSZ {J + cos2 y = I. Eq. 2-S.
FREE-BODY DIAGRAMS
185
186
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
TABLE 4-2 Supports for Rigid Bodies Subjected to Three-Dimensional Force Systems
Types of Connection
Reaction
Number of Unknowns
(1)
One unknown. The reaction is a force which acts away
from the member in the known direction of the cable.
cable
(2)
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
F
smooth surface support
(3)
One unknown. The reaction is a force which acts
perpendicular to the surface at the point of contact.
F
roller
(4)
Three unknowns. The reactions are three rectangular
force components.
ball and socket
(5)
Four unknowns. The reactions are two force and two
couple-moment components which act perpendicular to
the shaft. Note: The couple moments are generally not
applied if the body is supported elsewhere. See the
examples.
single journal bearing
continued
4.5
FREE·BODY DIAGRAMS
187
TABLE 4-2 Continued
Types of Connection
Reaction
Number of Unknowns
(6)
Five unknowns. The reactions are two force and three
couple-moment components. Note: The couple moments
are generally nor applied if the body is supported
elsewhere. See the examples.
single journal bearing
with square shart
(7)
Five unknowns. The reactions arc three force and two
couple-moment components. Note: The couple moments
are generally not applied if the body is supported
elsewhe re. Sec the exa mples.
single thrusl bearing
M
et
(8)
•
Fo//
~B:11
x
single smooth pin
Five unknowns. The reactions arc three force and two
couple-moment components. Note: The couple moments
are generally nor applied if the body is supported
elsewhere. See the examples.
(9)
single hinge
M,
Five unknowns. The reactions are three force and two
couple-moment components. Note: The couple moments
are generally nor applied if the body is supported
elsewhere. See the examples.
(10)
Six unknowns. The reactions arc three force and three
couple-moment components.
fixed support
188
CHAPTER
4
EQUILIBRIUM OF A RI GID BODY
Typical examples of actual supports that are referenced to Table 4-2 are
shown in the following sequence of photos.
The journal bearings support the ends
of the shaft. (5)
This ball-and-socket joint provides a
connection for a member of an earth
grader to its frame. (4)
This thrust bearing is used to support the
drive shaft on a machine. (7)
This pin is used to support the end of
the strut used on a tractor. (8)
Free-Body Diagrams. The general procedure for establishing the
free-body diagram of a rigid body has been outlined in Sec. 4.2. Essentially
it requires first "isolating" the body by drawing its outlined shape. This is
followed by a careful labeling of all the forces and couple moments with
reference to an established x, y, z coordinate system. As a general rule,
show the unknown components of reaction as acting on the free-body
diagram in the positive sense. In this way, if any negative values are
obtained, they will indicate that the components act in the negative
coordinate directions.
4.5
I EXAMPLE
189
FREE-BODY D IAGRAMS
4.9
Consider the two rods and plate, along with their associated free-body
diagrams, shown in Fig. 4-19. The x, y, z axes are established on each diagram
and the unknown reaction components are indicated in the positive sense.
The weight is neglected.
c
z
I
SOLUTION
•,u<:c,
<l"',
SOON
SOON
Properly aligned journal
bearings at A, B, C.
The force reactions developed by
the bearings are sufficient for
equilibrium since they prevent the
shaft from rotating about each of the
coordinate axes. No couple moments
at each bearing are developed.
z
I
M
A,cb
•
c
300 lb
B
Moment components are developed
by the pin on the rod to prevent
rotation about the x and z axes.
Pin at A and cable BC.
z
400 lb
A
B,
B
Properly aligned journal bearing
at A and hinge at C. Roller at B.
Only force reactions are developed by the
bearing and hinge on the plate to prevent
rotation about each coordinate axis. No
moments are developed at the hinge or bearing.
Fig. 4-19
190
CHAPTER
4
EQUILIBRIUM OF A RIG I D BODY
4. 6
EQUATIONS OF EQUILIBRIUM
As stated in Sec. 4.1, the conditions for equilibrium of a rigid body
subjected to a three-dimensional force system require that both the
resultant force and resultant couple moment acting on the body be equal
to zero.
Vector Equations of Equilibrium. The two conditions for
equilibrium of a rigid body may be expressed mathematically in vector
form as
I F= 0
I M0 = 0
(4-6)
where IF is the vector sum of all the external forces acting on the body,
and I M0 is the sum of the couple moments and the moments of all the
forces about any point 0 located either on or off the body.
Scalar Equations of Equilibrium. If all the external forces and
couple moments are expressed in Cartesian vector form and substituted
into Eqs. 4-6, we have
I F = If,i + IFyj + IFzk = 0
I Mo = I./\(,i + IMyj + IMzk = 0
Since the i,j , and k components are independent from one another, then
these equations are satisfied provided
IF.x = 0
IF.y = 0
IF.z = 0
(4-7a)
IMx = 0
IMy = 0
IMz = 0
(4-7b)
and
These six scalar equilibrium equations may be used to solve for at most
six unknowns shown on the free-body diagram. Equations 4-7a require
the sum of the external force components acting in the x, y, z directions
to be zero, and Eqs. 4-7b require the sum of the moment components
about the x, y, z axes to be zero.
4.6
IMPORTANT POINTS
• Always draw the free-body diagram first when solving any
equilibrium problem.
• If a support prevents tra11slation of a body, then the support
exerts a force on the body.
• If a support prevems rotation of a body, then the support exerts
a couple moment on the body.
PROCEDURE FOR ANALYSIS
Three-dimensional equilibrium problems for a rigid body can be
solved using the following procedure.
Free-Body Diagram.
• Draw an outlined shape of the body.
• Show all the forces and couple moments acting on the body.
• Establish the origin of the x, y, z axes at a convenient point and
orient the axes so that they are parallel to as many of the
external forces and moments as possible.
• Label all the loadings and specify their directions. In general,
show all the unknown components having a positive sense
along the x, y, z axes.
• Indicate the dimensions of the body necessary for calculating
the moments of forces.
Equations of Equilibrium.
• If the
x, y, z force and moment components seem easy to
determine, then apply the six scalar equations of equilibrium;
otherwise use the vector equations.
• It is not necessary that the set of axes chosen for force summation
coincide with the set of axes chosen for moment summation.
• Choose the direction of an axis for moment summation such
that it intersects the lines of action of as many unknown forces
as possible. Realize that the moments of forces passing throuigh
points on this axis, and the moments of forces which are
parallel to the axis, will then be zero.
• If the solution of the equilibrium equations yields a negative
scalar for a force or couple moment magnitude, it indicates that
the sense is opposite to that assumed on the free-body diagram.
EQUATIONS OF EOUIUBRIUM
1 91
192
CHAPTER
EXAMPLE
-
4
EQUILIBRIUM OF A RIG I D BODY
4.10
-
The homogeneous plate shown in Fig. 4-20a has a mass of 100 kg and is
subjected to a force and couple moment along its edges. If it is supported in
the horizontal plane by a roller at A, a ball-and-socket joint at B, and a cord
at C, determine the components of reaction at these supports.
300N200N·m
~
1.sn?'i
~
c
nf>
2m
SOLUTION (SCALAR ANALYSIS)
Free-Body Diagram. There are five unknown reactions acting on the
plate, as shown in Fig. 4-20b. Each of these reactions is assumed to act in
a positive coordinate direction.
~
B
(a)
z
300N
l 200N·m
J981 N~_JT ·
x_,.c;<'~~~(
f ><fw,P,1
' 1.5 m
A,
/
x' ,,._..
B,
By '....... ,
Y
B,
(b)
Fig. 4-20
Equations of Equilibrium. Since the three-dimensional geometry is
rather simple, a scalar analysis provides a direct solution to this problem.
A force summation along each axis yields
IF.x
- IF.y
O·,
= O·,
IFz = O·,
=
Bx
By
Ans.
Ans.
= 0
= 0
Az + Bz + Tc - 300 N - 981 N
=
(1)
0
Recall that the moment of a force about an axis is equal to the product
of the force magnitude and the perpendicular distance (moment arm)
from the line of action of the force to the axis. Also, forces that are
parallel to an axis or pass through it create no moment about the axis.
Hence, summing moments about the positive x and y axes, we have
IM., = O;
"I.My = O;
Tc (2 m) - 981 N(l m) + Bz(2 m)
=
0
-300 N(l.5 m) + 981 N(l.5 m) - Bz(3 m) - Az (3 m) - 200 N · m
(2)
0
(3)
The components of the force at B can be eliminated if moments are
summed about the x' and y' axes. We obtain
IM<' = O;
981 N(l m) + 300 N(2 m) - Az(2 m) = 0
(4)
"I.My· = O;
=
-300 N(l.5 m) - 981 N(l.5 m) - 200 N · m + Tc (3 m) = 0
(5)
Solving Eqs. (1) through (3) or the more convenient Eqs. (1), (4), and (5)
yields
Az = 790N Bz = -217N Tc = 707N
Ans.
The negative sign indicates that B z acts downward.
NOTE: The solution of this problem does not require a summation of
moments about the z axis. The plate is partially constrained because the
supports cannot prevent it from turning about the z axis if a force is
applied to it in the x- y plane.
4.6
EXAMPLE
4.11
-
-
Determine the components of reaction that the ball-and-socket joint at A,
the smooth journal bearing at B, and the roller support at C exert on the rod
assembly in Fig. 4-21a.
z
z
900N
900 N
(a)
(b)
Fig. 4-21
SOLUTION
As shown on the free-body diagram, Fig. 4-2lb,
the reactive forces of the supports will prevent the assembly from
rotating about each coordinate axis, and so the journal bearing at B only
exerts reactive forces on the member.
Free-Body Diagram.
A direct solution for A y can be obtained by
summing forces along they axis.
IFy = O;
Ay = 0
Ans.
The force Fe can be determined directly by summing moments about the
y axis.
IMy = O;
Fe(0.6 m) - 900 N(0.4 m) = 0
Fe = 600N
Ans.
Using this result, Bz can be determined by summing moments about the
x axis.
IMx = O·,
Bz(0.8 m) + 600 N(l.2 m) - 900 N(0.4 m) = 0
Bz = -450N
Ans.
The negative sign indicates that Bz acts downward. The force B x can be
found by summing moments about the z axis.
IMz = O;
-Bx(0.8 m) = 0 Bx = 0
Ans.
Thus,
IF.x = O·,
Ax = 0
Ans.
Finally, using the results of B z and Fe,
IFz = 0;
A z + ( -450 N) + 600 N - 900 N = 0
A z = 750N
Ans.
Equations of Equilibrium.
EQUATIONS OF EQUILIBRIUM
19 3
194
C H APT ER 4
EQU ILI BR I UM OF A RI GID BODY
PRELIMINARY PROBLEMS
P4-2.
Draw the free-body diagram of each object.
P4-3. In each case, write the moment equations about the
x, y, and z axes.
z
z
600N
300N
I
A
c,
B
300N
lm~
x
0.5 m
B,
B,
A,
(a)
(a)
z
B,
y
c,
SOON
(b)
(b)
z
I
z
B
A
A,
400N
1~~Y
....----~.,_~
m - -./
/ ~-,.
-~"'--- 2
c,
x
(c)
Prob. P4-2
(c)
Prob. P4-3
4.6
19 5
EQUATIONS OF EQUILIBRIUM
FUNDAMENTAL PROBLEMS
All problem solutions m ust include an FBD.
-·4-7. The uniform plate has a weight of 500 lb. Determine
the tension in each of the supporting cables.
14-1 • Determine the support reactions at the smooth
journal bearings A. 8, and C of the pipe assembly.
z
c
8
I
y
2001b
Prob. F4-IO
F4-ll. Determi ne the force developed in the short link
BD , and the tension in the cords CE and CF, and the
reactions of the ball-a nd-socket joi nt A on the block.
x
Prob. F4-7
F4-8. Determine the reactions at the roller support A, the
ball-and-socket joint D. and the tension in cable BC for the
plate.
l
I
l
cl
x
)'
0.1 m
6kN
Pr
4-9. The rod is supported by smooth journal bearings at
A, B, and C. Determjne the reactions at these supports.
9kN
}4-•
- 4-U. Determine the components of reaction that the
thrust bearing A and cable BC exert on the bar.
l
x
Prob. F<4-9
Prob. F4-12
196
C H APT ER
4
EQU ILI BR I UM OF A RI GID BODY
PROBLEMS
All problem solutions must include an FBD.
4-26. The uniform load has a mass of 600 kg and is lifted
using a uniform 30-kg strongback beam BAC and the four
ropes as shown. Determine the tension in each rope and the
force that must be applied at A.
*4-28. Determine the components of reaction at the fixed
support A. The 400 N, 500 N, and 600 N forces are parallel
to the x,y, and z axes, respectively.
z
F
1.25 m
600N
1.25 m
A
0.75m
A
SOON
x
y
0
Prob. 4-28
Prob.4-26
4-27. Due to an unequal distribution of fuel in the wing
tanks, the centers of gravity for the airplane fuselage A and
wings B and Care located as shown. If these components have
weights WA = 45 000 lb, W8 = 8000 lb, and We = 6000 lb,
determine the normal reactions of the wheels D, E, and Fon
the ground.
4-29. The 50-lb mulching machine has a center of gravity
at G. Determine the vertical reactions at the wheels C
and Band the smooth contact point A.
x
y
)'
Prob.4-27
Prob. 4-29
4.6
4-30. The smooth uniform rod AB is supported by a balland-socketjoint at A, the wall at B.and cable BC. Determine
the components of reaction at A, the tension in the cable,
and the normal reaction at 8 if the rod has a mass of 20 kg.
197
EQUATIONS OF EQUILIBRIUM
*4-32. The 100-lb door has its center of gravity at G.
Determine the components of reaction at hinges A and 8 if
hinge B resists only forces in the x and y directions and A
resists forces in the x, y, z directions.
18 in.
z
8
24 in.
24 in.
I
3D°
y
--<_y
x
x
Prob.4-32
Prob. 4-30
4-33. Determine the tension in each cable and the
components of reaction at D needed to support the load.
z
4-31. The uniform concrete slab has a mass of 2400 kg.
Determine the tension in each of the three parallel
supporting cables.
z
I
x- A
15 kN
y
Tc
A
30°
x
0.5 111
400N
Prob. 4-31
Prob.4-33
198
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
4-34. The bent rod is supported at A , B, and C by smooth
journal bearings. Calculate the x, y, z components of reaction
at the bearings if the rod is subjected to forces Fi = 300 lb
and Fi = 250 lb. F 1 lies in the y-z plane. The bearings are in
proper alignment and exert only force reactions on the rod.
*4-36. The bar AB is supported by two smooth collars.
At A the connection is a ball-and-socket joint and at Bit is
a rigid attachment. If a 50-lb load is applied to the bar,
determine the x, y, z components of reaction at A and B.
1 ft
A
c
-3ft-
30"
y
x
Prob. 4-34
Prob.4-36
4-35. The bent rod is supported at A , B, and C by smooth
journal bearings. Determine the magnitude of F2 which will
cause the reaction c,. at the bearing C to be equal to zero.
The bearings are in proper alignment and exert only force
reactions on the rod. Set F1 = 300 lb.
4-37. The rod has a weight of 6 lb/ft. If it is supported by a
ball-and-socket joint at C and a journal bearing at D ,
determine the x, y, z components of reaction at these
supports and the moment M that must be applied along the
axis of the rod to hold it in the position shown.
z
c
-3ft-
30"
y
x
y
Prob. 4-35
Prob.4-37
4.6
4-38. The sign has a mass or 100 kg with center of mass
at G. Determine the x, y, z components of reaction at the
ball-and-socket joint A and the tension in wires BC and BD.
199
EQUATIONS OF EOUIUBRIUM
*4-40. Both pulleys are fixed to the shaft and as the shaft
turns with constant angular velocity, the power of pulley A
is transmitted to pulley B. Determine the horizontal tension
T in the belt on pulley B and the x. y, z components of
reaction at the journal bearing C and thrust bearing D if
0 = 45°. The bearings are in proper alignment and exert
only force reactions on the shaft.
z
200m~
I
·~~~:pi--__....,-
250 m~
300m~
r---.....y
0
~_s,C~
x~
•G
-lm-k
lm-J
65N
l
T
SON
Prob. 4-38
Prob. 4-40
4-39. Both pulleys are fixed to the shaft and as the shaft
turns with constant angular velocity, the power of pulley A
is transmitted to pulley B. Determine the horizontal tension
T in the belt on pulley B and the x, y. z components of
reaction at the journal bearing C and thrust bearing D if
0 = O". The bearings are in proper alignment and exert only
force reactions on the shaft.
4-41. Member AB is supported by a cable BC and at A by
a square rod which fits loosely through the square hole
at the end collar of the member as shown. Determine the
x, y, z components of reaction at A and the tension in the
cable needed 10 hold the 800-lb cylinder in equilibrium.
z
200m~.~~~l~Jl'-__,-50N
250m~
300m~
&
~
B
3 ft>---
x
l
'<:.:'._ y
T
65 N
SON
Prob. 4-39
50N
Prob.4-41
200
CHAPTER 4
EQU I LIBR I UM OF A R I GI D BODY
4.7
w
•
p
-
(a)
w
CHARACTERISTICS OF DRY
FRICTION
Friction is a force that resists the movement between two contacting
surfaces that slide relative to one another. This force always acts tangent
to the surface at the points of contact, and is directed so as to oppose the
possible or existing motion between the surfaces.
In this section, we will study the effects of dry f riction, which is
sometimes called Coulomb friction since its characteristics were studied
extensively by C.A. Coulomb in 1781. Dry friction occurs between the
contacting surfaces of bodies when there is no lubricating fluid.
Theory of Dry Friction. The theory of dry friction can be explained
p
•
(b)
(c)
by considering the effects caused by pulling horizontally on a block of
uniform weight W which is resting on a rough horizontal surface,
Fig. 4- 22a. As shown on the free-body diagram of the block, Fig. 4-22b,
the floor exerts an uneven distribution of both normal force d N,, and
frictional force d F,, along the contacting surface.* For equilibrium, the
normal forces must act upward to balance the block's weight W, and the
frictional forces act to the left to prevent the applied force P from moving
the block to the right. Close examination of the contacting surfaces
between the floor and block reveals how these frictional and normal
forces develop, Fig. 4-22c. It can be seen that many microscopic
irregularities exist between the two surfaces and, as a result, reactive
forces d R,, are developed at each point of contact. As shown, each
reactive force contributes both a frictional component d F,, and a normal
component d N,,.
Equilibrium. The effect of the distributed normal and frictional
w
1-a/2 ; 2-1
p
•
h
f--1--,....,.~ F
x N
Resultant Nom1al
and Frictional Forces
(d)
loadings is indicated by their resullllnts N and F on the free-body diagram
shown in Fig. 4-22d. Notice that N acts a distance x to the right of the line
of action of W, Fig. 4-22d. This location of the normal force distribution
in Fig. 4-22b is necessary in order to balance the "tipping effect" caused
by P. For example, if P is applied at a height h from the surface,
Fig. 4-22d, then moment equilibrium about point 0 is satisfied if
Wx = Phorx = Ph/W.
Fig. 4-22
* A complete discussion of distributed loadings is given in Sec. 3.9.
4. 7
-- ~
P
CHARACTERISTICS OF
DRY
FRICTION
201
Impending
.
mollon
(e)
Fig. 4-22 (cont.)
Impending Motion. In cases where the surfaces of contact are
rather "slippery," the frictional force F may not be great enough to
balance P, and consequently the block will tend to slip. In other words, as
P is slowly increased, F correspondingly increases until it attains a certain
maximum value F,, called the limiting static frictional force, Fig. 4-22e.
When this value is reached, the block is in unstable equilibrium since any
further increase in P will cause the block to move. Experimentally, it has
been determined that F, is directly proportional to the resultant normal
force N. Expressed mathematically,
(4-8)
where the constant of proportionality, I.Ls (mu "sub" s), is called the
coefficient of static f riction.
Thus, when the block is on the verge ofsliding, the normal force N and
frictional force F, combine to create a resultant Rs, Fig. 4-22e. The angle
<Ps (phi) that ~ makes with N is called the angle of static friction. From
the figure,
[Tllble4-3
!YPk* v.i.... for
Contact
Materials
Typical values for µ,, are given in Table 4-3. As indicated, these values
will vary since experimental testing was done under variable conditions
of roughness and cleanliness of the contacting surfaces. For applications,
therefore, it is important that both caution and judgment be exercised
when selecting a coefficient of friction for a given set of conditions.
When a more accurate calculation of F, is required, the coefficient of
friction should be determined directly by an experiment that involves
the two contacting materials.
Coefficient of
Static Friction (µ,)
Metal on ice
0.03-0.05
Wood on wood
0.30-0.70
Lea ther on wood
0.20-0.50
Leather on metal
0.30-0.60
Copper on copper
0.74-1.21
I
202
C H APT ER 4
EQU ILI BR I UM OF A RI GI D BODY
w
~~1-~ -
M o tion
p
•
(a)
Fig. 4-23
Motion. If the magnitude of P acting on the block is increased so that
it becomes slightly greater than F,, the frictional force at the contacting
surface will drop to a smaller value Fh called the kinetic frictional force.
The block will then begin to slide with increasing speed, Fig. 4- 23a. As
this occurs, the block will " ride" on top of the peaks at the points
of contact, as shown in Fig. 4-23b. The continued breakdown of the
nonrigid surfaces is the dominant mechanism creating kinetic friction.
Experiments with sliding blocks indicate that the magnitude of the
kinetic friction force is directly proportional to the magnitude of the
resultant normal force, expressed mathematically as
(4-9)
Here the constant of proportionality, /.Lh is called the coefficient of
kinetic friction . Typical values for /.Lk are approximately 25 percent
smaller than those listed in Table 4-3 for JLs·
As shown in Fig. 4- 23a, in this case, the resultant force at the surface of
contact, Rh has a line of action defined by <Pk· This angle is referred to as
the angle of kinetic friction , where
"'
't'k
(Fk)
_(µ,kN)
N = tan N
_1
= tan
By comparison, <Ps =::: <Pk·
1
= tan
_
1
/.Lk
4.7
CHARACTERISTICS OF
Characteristics of Dry Friction. As a result of experiments that
pertain to the foregoing discussion, we can state the following rules
which apply to bodies subjected to dry friction.
•
The frictional force acts tangent to the contacting surfaces in a
direction opposed to the motion or tendency for motion of one
surface relative to another.
•
The maximum static frictional force F, that can be developed is
independent of the area of contact between the surfaces, provided
the normal pressure is not very low nor great enough to severely
deform or crush the surfaces between the bodies.
•
The maximum static frictional force is generally greater than the
kinetic frictional force for any two surfaces of contact. However, if
one of the bodies is moving with a very low velocity over the surface
of another, Fk becomes approximately equal to F,, i.e., f.Ls "" f-lk·
•
When slipping at the surface of contact is about to occur, the
maximum static frictional force is proportional to the normal force,
such that F, = µ,5 N.
•
When slipping at the surface of contact is occurring, the !kinetic
frictional force is proportional to the normal force, such that
Fk = f.LkN.
w
T
F
Some objects, such as this barrel, may not be on the verge of slipping,
and therefore the friction force F must be determined strictly from
the equations of equilibrium.
DRY FRICTION
203
204
CHAPTER
4
EQUILIBRIUM OF A RIG I D BODY
4. 8
PROBLEMS INVOLVING DRY
FRICTION
If a rigid body is in equilibrium when it is subjected to a system of
forces that includes the effect of friction, the force system must satisfy not
only the equations of equilibrium but also the Jaws that govern the
frictional forces.
A
!LA = 0.3
(a)
Types of Friction Problems. In general, there are three types of
static problems involving dry friction. They can easily be classified once
free-body diagrams are drawn and the total number of unknowns are
identified and compared with the total number of available equilibrium
equations.
(b)
Fig. 4-24
B
A
\8
JLA = Q.3
JLB = 0.4
No Apparent Impending Motion. Problems in this category are
strictly equilibrium problems, which require the number of unknowns to
be equal to the number of available equilibrium equations. Once the
frictional forces are determined from the solution, however, their
numerical values must be checked to be sure they satisfy the inequality
F ::;; µ, N ; otherwise, slipping will occur and the body will not remain in
equilibrium. A problem of this type is shown in Fig. 4-24a. Here we must
determine the frictional forces at A and C to check if the equilibrium
position of the two-member frame can be maintained. If the members
are uniform and have known weights of 100 N each, then the free-body
diagrams are as shown in Fig. 4-24b. There are six unknown force
components which can be determined strictly from the six equilibrium
equations (three for each member). Once FA, NA, Fe, and Ne are
determined, then the members will remain in equilibrium provided
FA ::;; 0.3NA and Fe ::;; 0.5Ne are satisfied.
(a)
(b)
Fig. 4-25
Impending Motion at All Points of Contact. In this case the total
number of unknowns will equal the total number of available equilibrium
equations plus the total number of available frictional equations, F = µN.
When motion is impending at the points of contact, then F, = µ, N; whereas
if the body is slipping, then Fk = /Lk N. For example, consider the problem of
finding the smallest angle 8 at which the 100-N bar in Fig. 4-25a can be
placed against the wall without slipping. The free-body diagram of the bar is
shown in Fig. 4-25b. Here the five unknowns are determined from the three
equilibrium equations and two static frictional equations which apply at
both points of contact, so that FA = 0.3NA and F8 = 0.4N8 .
4.8
PROBLEMS INVOLVING
Impending Motion at Some Points of Contact. For these types of
problems, the number of unknowns will be less than the number of
available equilibrium equations plus the number of available frictional
equations or conditional equations for tipping. As a result, several
possibilities for motion or impending motion will exist and the problem
will involve a determination of the kind of motion which actually occurs.
For example, consider the two-member frame in Fig. 4-26a. In this
problem we wish to determine the horizontal force P needed to cause
movement. If each member has a weight of 100 N, then the free-body
diagrams are as shown in Fig. 4-26b. There are seven unknowns. For a
unique solution we must satisfy the si.x equilibrium equations {three for
each member) and only one of two possible static frictional equations.
This means that as P increases it will either cause slipping at A and no
slipping at C, so that FA = 0.3NA and Fe :s 0.5Ne, or slipping wiU occur
at C and no slipping at A , in which case Fe = 0.5Ne and FA < 0.3NAThe actual situation can be determined by calculating P for each case,
and then choosing the case for which P is smaller. If in both cases the
same value for P is calculated, which would be highly improbable, then
slipping at both points occurs simultaneously; i.e., the seven unknowns
would have to satisfy eight equations.
DRY
B
A
µ.,. ~
""c
0.3
JJ.c
(a)
(b)
Equilibrium Versus Frictional Equations. Whenever we solve
a problem such as the one in Fig. 4-24, where the friction force Fis to be
an "equilibrium force" and satisfies the inequality F < µ,,N, then we can
assume the sense of direction of Fon the free-body diagram. The correct
sense is made known after solving the equations of equilibrium for F. If F
is a negative scalar the sense of F is the reverse of that which was
assumed. This convenience of assuming the sense of F is possible because
the equilibrium equations equate to zero the components of vectors
acting in the same direction. H owever, in cases where the frictional
equation F = µ.N is used in the solution of a problem, this convenience
of assuming the sense of F is lost, since the frictional equation relates
only the magnitudes of two perpendicular vectors. Consequently, !F must
always be shown acting with its correct sense on the free-body diagram,
whenever the frictional equation is used for the solution of a problem.
205
FRICTION
Fig. 4-26
= 0.5
206
CHAPTER 4
EQUILIBRIUM OF A RIG I D BODY
IMPORTANT POINTS
• Friction is a tangential iforce that resists the movement of one
surface relative to another.
• If no sliding occurs, the maximum value for the friction force is
equal to the product of the coefficient of static friction and the
normal force at the surface, F, = /Ls N.
• If sliding occurs, then the friction force is the product of the
coefficient of kinetic friction and the normal force at the surface,
Fk = ILkN.
• There are three types of static friction problems. Each of these
problems is analyzed by first drawing the necessary free-body
diagrams, and then applying the equations of equilibrium,
while satisfying either the conditions of friction or the
possibility of tipping.
1-b/2- >-b/2-1
p
1-b/2- -b/2-1
p
,,
!w
!w
•
•
h
-x
F
F
,_
N
- x-
Consider pushing on the uniform crate that has a weight Wand sits on the rough surface. As shown on the first
free-body diagram, if the magnitude of P is small, the crate will remain in equilibrium. As P increases the crate will
either be on the verge of slipping on the surface (F = µ,, N), or if the surface is very rough (largeµ,,), then the
resultant normal force will shift to the corner, x = b /2, as shown on the second free-body diagram. At this point
the crate will begin to tip over. The crate also has a greater chance of tipping if P is applied at a greater height h
above the surface, or if its width b is smaller.
4.8
PROBLEMS INVOLVING
PROCEDURE FOR ANALYSIS
Equilibrium problems involving dry friction can be solved using the
following procedure.
Free-Body Diagrams.
• Draw the necessary free-body diagrams, and unless it is stated
in the problem that impending motion or slipping occl!lrs,
always show the frictional forces as unknowns (i.e., do not
assume F = µ,N).
• Determine the number of unknowns and compare this with the
number of available equilibrium equations.
• If there are more unknowns than equations of equilibrium, it will
be necessary to apply the frictional equation at some, if not all,
points of contact to obtain the extra equations needed for a
complete solution.
• If the equation F = µ,N is to be used, it will be necessary to show
F acting in the correct sense of direction on the free-body
diagram.
Equations of Equilibrium and Friction.
• Apply the equations of equilibrium and the necessary frictional
equations (or conditional equations if tipping is possible) and
solve for the unknowns.
• If the problem involves a three-dimensional force system such
that it becomes difficult to obtain the force components or the
necessary moment arms, apply the equations of equilibrium
using Cartesian vectors.
DRY FRICTION
207
208
I
CHAPTER 4
EXAMPLE
EQUILIBRIUM OF A RIG I D BODY
4. 12
I
The uniform crate shown in Fig. 4-27a has a mass of 20 kg. If a force
P = 80 N is applied to the crate, determine if it remains in equilibrium.
The coefficient of static friction is /Ls = 0.3.
1 - - -0.8 01 - - - 1
P= 80N
(a)
Fig. 4-27
SOLUTION
As shown in Fig. 4-27b, the resultant normal
force Ne must act a distance .x from the crate's centerline in order to
counteract the tipping effect caused by P. There are three unknowns,
F, N e, and x, which can be determined strictly from the three equations
of equilibrium.
Free-Body Diagram.
P= 80N
- x-
Equations of Equilibrium.
~IF, = O;
Ne
(b)
80 cos 30° N - F
=
0
+ jlF.y
=
O·,
C+lM0
=
O; 80 sin 30°N(0.4 m)-80cos 30°N(0.2 m) + Ne(x)
-80sin 30°N + Ne - 196.2 N
=
0
=
0
Solving,
F
x =
=
69.3 N
Ne = 236.2N
-0.00908 m = -9.08 mm
Since xis negative it indicates the resultant normal force acts (slightly)
to the left of the crate's centerline. Also, the maximum frictional force
which can be developed at the surface of contact is
Fmax = µ,5 N e = 0.3(236.2 N) = 70.9 N. Since F = 69.3 N < 70.9 N,
the crate will not slip, although it is very close to doing so.
4.8
EXAMPLE
PROBLEMS INVOLVING
DRY
F RICTION
4 .13
It is observed tha t when the bed of the dump truck is raised to an angle o f
0 = 25° the vending machines will begin to slide off the bed, Fig. 4-2&z.
Determine the coefficient of static friction between a vending machine
and the su rface of the truckbed .
SOLUTION
An idealized m od el of a ve nding machine resting on the truckbe d is
shown in Fig. 4-28b. The dimensions have been measured and the
ce nte r o f gravity has been located. We will assume that the vending
machine we ighs W.
(a)
Free-Body Diagram. As shown in Fig. 4-28c, the dimension x is used
to locate the position of the resultant normal force N . There ar e four
unknowns, N, F, µ,, , and x.
Equations of Equilibrium.
= O·'
+ ?lF,. = O;
C+ lM0 = O;
+\.l F..r
W sin 25° - F = 0
(1)
=0
+ W cos 25°(x) = O
(2)
N - W cos25°
- W sin 25°(2.5 ft)
(b)
(3)
Since slipping im pends at 0 = 25°, using Eqs. 1 and 2, we have
F,
= µ,, N;
W sin 25° = µ,, ( W cos 25°)
µ,,
= tan 25° = 0.466
Ans.
The angle of O = 25° is referred to as the angle of repose,and by comparison,
it is equal to the angle of static friction, 0 = <Ps· Since 0 is independent of
the weight of the vending machine, knowing 0 provides a convenient
method for determining the coefficient of static friction.
NOTE: Fro m E q. 3, we find
x
= 1.17 ft. Since 1.17 ft <
1.5 ft, indeed
the ve nding machine will slip down the truckbed and not tip over.
(c)
Fig. 4-28
209
210
I
CHAPTER
EXAMPLE
4
4. 14
EQUILIBRIUM OF A RIG I D BODY
I
Blocks A and B have a mass of 3 kg and 9 kg, respectively, and are
connected to the weightless links shown in Fig. 4-29a. Determine the
largest vertical force P that can be applied to the pin C without causing
any movement. The coefficient of static friction between the blocks and
the contacting surfaces isµ,, = 0.3.
P
SOLUTION
Free-Body Diagram. The links are two-force members and so the free(a)
body diagrams of pin C and blocks A and Bare shown in Fig. 4-29b. Since
the horizontal component of FAc tends to move block A to the left, FA
must act to the right. Similarly, F8 must act to the left to oppose the
tendency of motion of block B to the right,caused by FBC· There are seven
unknowns and six available force equilibrium equations, two for the pin
and two for each block, so that only one friction equation is needed.
Equations of Equilibrium and Friction. The force in links AC and
BC can be related to P by considering the equilibrium of pin C.
p
+ jIF.y
=
O·,
FAccos 30° - P
~IF...t = O·'
- -04 - - - - -x
C
Fsc
=
FAc
O;
l.l55P sin 30° - F8 c = O;
=
1.155P
F8 c = 0.5774P
Using the result for FA c, for block A,
~IF, = O;
+f
IFy
=
O;
FA - 1.155P sin 30° = O; FA = 0.5774P
NA -1.155P cos 30° - 3(9.81 N) = O;
NA = P + 29.43N
Using the result for F8 c, for block B ,
~IF, = O;
(0.5774P) - F8 = O;
F8
+f
IFy
=
0;
N8
-
9(9.81) N
=
(1)
(2)
=
0.5774P
(3)
N 8 = 88.29N
O;
Movement of the system may be caused by the initial slipping of eilher
block A or block B. If we assume that block A slips first, then
9(9.81) N
'"=0577~
~
Ns
(b)
Fig. 4-29
FA
=
/.LsNA
=
0.3NA
(4)
Substituting Eqs. 1 and 2 into Eq. 4,
0.5774P
P
= 0.3(P
=
+ 29.43)
31.8N
Ans.
Substituting this result into Eq. 3, we obtain F8 = 18.4 N. Since the
maximum static frictional force at B is (Fs)max = /.LsNB =
0.3(88.29 N) = 26.5 N > F8 , block B will not slip. Thus, the above
assumption is correct. Notice that if the inequality were not satisfied,
we would have to assume slipping of block B (F8 = 0.3 N8 ) and then
solve for P.
4.8
PROBLEMS INVOLVING
DRY
211
FRICTION
PRELIMINARY PROBLEMS
P4-4. Detennine the friction force at the surface of contact.
SOON
P4-6. Detennine the force P to move block B.
J
~
IW=200N
I
W= 100 N
A
v
µ, = 0.2 .
W= 100 N
p
B
0.3
1-'k = 0.2
µ, =
µ, = 0.2
v
c
(a)
µ, -
IOON
~
W=200 N
0.1
Prob. P4-6
IW=40N
I
= 0.9
1-'k = 0.6
µ,
(b)
P4-7. Determine the force P needed to cause impending
motion of the block.
Prob. P4-4
p
P4-5. Detennine the couple moment M needed to cause
impending motion of the cylinder.
-
I
2m
W=200N
I
f-tm-1
µ, - 0.3
(a)
W= IOON .---...._
B
Smooth
T
Im
l Iµ,, = 0.1
Prob. P4-5
W - lOON
I 111 -
Iµ, = 0.4
(b)
Prob. P4-7
212
C H APT ER
4
EQU ILI BR I UM OF A RI GID BODY
FUNDAMENTAL PROBLEMS
All problem solutions must include FBDs.
F4-13. Determine the friction developed between the 50-kg
crate and the ground if a) P = 200 N, and b) P = 400 N.
The coefficients of static and kinetic friction between the crate
and the ground areµ., = 0.3 and µ.k = 0.2.
F4-16. If the coefficient of static friction at contact points A
and B is µ., = 0.3, determine the maximum force P that can
be applied without causing the 100-kg spool to move.
Prob. F4-13
F4-14. Determine the minimum force P to prevent the
30-kg rod AB from sliding. The contact surface at B is
smooth, whereas the coefficient of static friction between
the rod and the wall at A is µ., = 0.2.
Prob. F4-16
A
1
l
3m
B
p
F4-17. Determine the maximum force P that can be
applied without causing movement of the 250-lb crate that
has a center of gravity at G. The coefficient of static friction
at the floor isµ., = 0.4.
i - - - - - 4 m - - - -1
Prob. F4-14
F4-15. Determine the maximum force P that can
be applied without causing the two 50-kg crates to move.
The coefficient of static friction between each crate and the
ground isµ., = 0.25.
[I
A
B
~ ~:1
Prob. F4-15
I~
Prob. F4-17
4.8
F4-18. Determine the m1mmum coefficient of static
friction between the uniform 50-kg spool and the wall so
that the spool does not slip.
PROBLEMS INVOLVING
DRY FRICTION
213
F4-20. If the coefficient of static friction at all contacting
surfaces is µ,,, determine the inclination 8 at which the
identical blocks, each of weight W, begin to slide.
6(f
0.6m
0.3m
Prob. F4-20
Prob. F4-18
F4-19. Blocks A , B, and C have weights of 50 N, 25 N, and
15 N, respectively. Determine the smallest horizontal force P
that will cause impending motion. The coefficient of static
friction between A and B is J.Ls = 0.3, between B and C,
µ!, = 0.4, and between block C and the ground,µ!,' = 0.35.
'
300mm
-,
A
p
B
-
F4-21. Blocks A and B have a mass of 7 kg and 10 kg,
respectively. Using the coefficients of static friction
indicated, determine the largest force P which can be
applied to the cord without causing motion. There are
pulleys at C and D.
-
1
1
DO
8
c.,.0....,~--~P
c
J
D
Prob. F4-19
J.LA = Q.1
Prob. F4-21
214
C H APT ER 4
EQU ILI BR I UM OF A RI GID BODY
PROBLEMS
A ll problem solutions must include FBDs.
4-42. Determine the maximum force P the connection
can support so that no slipping occurs between the plates.
There are four bolts used for the connection and each is
tightened so that it is subjected to a tension of 4 kN. The
coefficient of static friction between the plates isµ., = 0.4.
*4-44. The mine car and its contents have a total mass of
6 Mg and a center of gravity at G. If the coefficient of static
friction between the wheels and the tracks is µ., = 0.4 when
the wheels are locked, find the normal force acting on the
front wheels at B and the rear wheels at A when the brakes
at both A and Bare locked. Does the car move?
lOkN
.c
0.9m
Prob. 4-42
- 0.6m 1 - - -1.S m- -- 1
4-43. The tractor exerts a towing force T = 400 lb.
Determine the normal reactions at each of the two front
and two rear tires and the tractive frictional force F on each
rear tire needed to pull the load forward at constant velocity.
The tractor has a weight of 7500 lb and a center of gravity
located at Gr. An additional weight of 600 lb is added to its
front having a center of gravity at GA- Take µ., = 0.4.
The front wheels are free to roll.
2.5 fl
- 4ft-
Prob. 4-43
3 ft
Prob. 4-44
4-45. The winch on the truck is used to hoist the garbage
bin onto the bed of the truck. If the loaded bin has a weight
of 8500 lb and center of gravity at G, determine the force in
the cable needed to begin the lift. The coefficients of static
friction at A and B are J.l.A = 0.3 and J.l.B = 0.2, respectively.
Neglect the height of the support at A.
--
.G
Al -10 ft-Jc---
Prob. 4-45
4.8
4-46. The automobile has a mass of 2 Mg and center of
mass al G. Determine the towing force F required to move
the car if the back brakes are locked, and the front wheels
are free to roll. Takeµ, = 0.3.
~7.
The automobile has a mass of 2 Mg and center of
mass at G. Determine the towing force F required to move
the car. Both the front and rear brakes are locked.
Takeµ, = 0.3.
PROBLEMS INVOLVING
DRY
215
FRICTION
~9.
The block brake consists of a pin-connected lever
and friction block at B. The coefficient of static friction
between the wheel and the lever isµ., = 0.3, and a torque of
5 N · m is applied 10 the wheel. Determine if the brake can
hold the wheel stationary when the force applied to the
lever is (a) P = 30 N. (b) P = 70 N.
~
150 mn1
f
p
A
1----1-Lm
Prob. ~9
0.75 111
Probs. 4-46/47
4-50. The pipe of weight Wis to be pulled up the inclined
plane of slope a using a force P. lf P acts at an angle .p, show
that for slipping P = W sin(a + 9)/cos(<P - 8), where 8 is
the ang.le of static friction; 8 = tan- 1 µ., .
·~8.
The block brake consists of a pin-connected lever
and friction block at B. The coefficient of static friction
between the wheel and the lever isµ, = 0.3. and a torque of
5 N · m is applied to the wheel. Determine if the brake can
hold the wheel stationary when the force applied to the
lever is (a) P = 30 N. (b) P = 70 N.
4-5L Determine the angle <P at which the applied force P
should act on the pipe so that the magnitude of P is as small
as possible for pulling the pipe up the incline. What is the
corresponding value of P? The pipe weighs Wand the slope
a is known. Express the answer in terms of the angle of
kinetic friction, 8 = tan- 1 µ. 4 •
~5N·m
~
150 mm
p
200 mm
400 mm ----1
Prob. 4-48
Probs. 4-50/51
216
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
*4-52. The log has a coefficient of static friction of
J.Ls = 0.3 with the ground and a weight of 40 lb /ft. If a man
can pull on the rope with a maximum force of 80 lb,
determine the greatest length I of log he can drag.
( iO
.. 80 lb
4-55. The spool of wire having a weight of 300 lb rests on
the ground at B and against the wall at A. Determine the
force P required to begin pulling the wire horizontally off
the spool. The coefficient of static friction between the
spool and its points of contact is J.Ls = 0.25.
*4-56. The spool of wire having a weight of 300 lb rests on
the ground at B and against the wall at A. Determine the
normal force acting on the spool at A if P = 300 lb.
The coefficient of static friction between the spool and the
ground at B is J.Ls = 0.35. The wall at A is smooth.
A
Prob. 4-52
3 ft
4-53. The 180-lb man climbs up the ladder and stops at the
position shown after he senses that the ladder is on the verge
of slipping. Determine the inclination e of the ladder if the
coefficient of static friction between the friction pad A and the
ground is J.Ls = 0.4. Assume the wall at B is smooth. The center
of gravity for the man is at G. Neglect the weight of the ladder.
4-54. The 180-lb man climbs up the ladder and stops at the
position shown after he senses that the ladder is on the verge
of slipping. Determine the coefficient of static friction between
the friction pad at A and ground if the inclination of the ladder
is e = 60° and the wall at B is smooth. The center of gravity
for the man is at G. Neglect the weight of the ladder.
0
1 ft
A
, /'!'....... -... p
Probs. 4-55/56
4-57. The ring has a mass of 0.5 kg and is resting on the
surface of the table. To move the ring a normal force P from
the finger is exerted on it as shown. Determine its magnitude
when the ring is on the verge of slipping at A. The coefficient
of static friction at A is J.LA = 0.2 and at B , µ,8 = 0.3.
A
Probs. 4-53154
Prob. 4-57
4.8
4-58. Determine the smallest force P that must be applied
in order to cause the 150-lb uniform crate to move. The
coefficent of static Criction between the crate and the floor
isµ., = 0.5.
4-59. The man having a weight of 200 lb pushes
horizontally on the crate. If the coefficient of static friction
between the 450-lb crate and the floor is µ., = 03 and
between his shoes and the floor isµ~ = 0.6, determine if he
can move the crate.
217
PROBLEMS INVOLVING DRY FRICTION
4-62. Determine the minimum force P needed to push
the tube E up the incline. The coefficients of static friction
at the contacting surfaces are JLA = 0.2, µ 8 = 0.3, and
JLc = 0.4. The 100-kg roller and 40-kg tube each have a
radius of 150 mm.
j-2rt-I
c
Prob. 4-62
••
Probs. 4-58/59
*4-60. The uniform hoop of weight W is subjected to the
horizontal force P. Determine the coefficient of static
friction between the hoop and the surface al A and B if the
hoop is on the verge of rotating.
4-61. Determine the maximum horizontal force P that
can be applied to the 30-lb hoop without causing it to rotate.
The coefficient of static friction between the hoop and the
surfaces A and B isµ, = 0.2. Take r = 300 mm.
4-63. The coefficients of static and kinetic friction
between the drum and brake bar arcµ" = 0.4 and /.Lk = 0.3,
respectively. lf M = 50 N · m and P = 85 N, determine the
horizontal and vertical components of reaction at the pin 0.
Neglect the weight and thickness of the brake. The drum has
a mass of 25 kg.
*4-64. The coefficient of static friction between the drum
and brake bar is µ., = 0.4. If the moment M = 35 N • m.
determine the smallest force P that needs to be applied to
the brake bar in order to prevent the drum from rotating.
Also determine the corresponding horizontal and vertical
components of reaction at pin 0. Neglect the weight and
thickness of the brake bar. The drum has a mass of 25 kg.
I 300 mm I
!
Olm
700 mm - --1
fo!A
p
500mm
A'•
A
Probs.~/61
1
~ 12Smm
Probs. 4-63/64
l
218
CHAPTER
4
EQUILIBRIUM OF A RIG ID BODY
CHAPTER REVIEW
Equilibrium
z
~
i..
A body in equilibrium is at rest or moves
with constant velocity.
FJ X
0
x
Two Dimensions
Before analyzing the equilibrium of a body,
it is first necessary to draw its free-body
diagram. This is an outlined shape of the
body, which shows all the forces and couple
moments that act on it.
Couple moments can be placed anywhere
on the free-body diagram since they are
free vectors. Forces can act at any point
along their line of action since they are
sliding vectors.
Angles used to resolve forces, and
dimensions used to take moments of the
forces, should also be shown on the freebody diagram.
Remember that a support will exert a
force on the body in a particular direction
if it prevents translation of the body in
that direction, and it will exert a couple
moment on the body if it prevents
rotation.
The three scalar equations of equilibrium
can be applied when solving problems in
two dimensions, since the geometry is
easy to visualize.
For the most direct solution, try to sum
forces along an axis that will eliminate as
many unknown forces as possible. Sum
moments about a point A that passes
through the line of action of as many
unknown forces as possible.
~)-,_-- 2m
~OON·~1
~~i:====1~~:::1= 11
,.A
_L B~(f '
1-- 2m -~OON·m
A, 4
--I ~Fsc
+
Ay
"
1m
_I_
'L x
=0
kF,. = 0
kF,
~M0 =
0
/
F1
Fi
\ 3(f
=-----i::.
/
-
F•
)'
219
CHAPTER REVIEW
Three Dimensions
In three dimensions, it is often advantageous
to use a Cartesian vector analysis when
applying the equations of equilibrium. To
do this, first express each known and
unknown force and couple moment shown
on the free-body diagram as a Cartesian
vector. Then set the force summation equal
to zero. Take moments about a point 0 that
lies on the line of action of as many
unknown force components as possible.
From point 0 direct position vectors to
each force, and then use the cross product
to determine the moment of each force.
The six scalar equations of equilibrium
are established by setting the respective i,
j , and k components of these force and
moment summations equal to zero.
LF = 0
LM = 0
·'
LMy = 0
LM- = 0
-
Dry Friction
Frictional forces exist between two rough
surfaces of contact. These forces act on a
body so as to oppose its motion or
tendency of motion.
A static frictional force has a maximum
val ue of F, = µ,N, where µ, is the
coefficienr of sraric friction. In this case,
motion between the contacting surfaces is
impending.
If slipping occurs, then the friction force
remains essentially constant and equal to
Fk = µkN. Here µk is the coefficieni of
kineric friciion.
The solution of a problem involving
friction requires first drawing the
free-body diagram of the body. If the
unknowns cannot be determined strictly
from the equations of equilibrium, and
the possibility of slipping occurs, then the
friction equation should be applied at the
appropriate points of contact in order to
complete the solution.
w
w
p
'
p
-
F
Rough surface
N
w
'
•
1---~p
.,.._1--r-N--
Impending
·- - - -+
motion
F, = µ,, N
w
'
Motion
1---.~p - - - -
N
220
CHAPTER 4
EQUILIBRIUM OF A RIG ID BODY
REVIEW PROBLEMS
All problem solutions must include an FBD.
R4-1. If the roller at B can sustain a maximum load of
3 kN, determine the largest magnitude of each of the three
forces F that can be supported by the truss.
R4-3. Determine the normal reaction at the roller A and
horizontal and vertical components at pin B for equilibrium
of the member.
lO kN
1 -0.6m- -0.6m- J
-
2m - - 2m
-
F
F
F
2m
Prob. R4-1
Prob. R4-3
R4-2. Determine the reactions at the supports A and B
for equilibrium of the beam.
*R4-4. Determine the horizontal and vertical components
of reaction at the pin A and the reaction at the roller B on
the lever.
400 N/m
~-----
2OON/m
'
.----
I
B
!lB
1 - -20 in.- - -1
Prob. R4-2
Prob. R4-4
221
REVIEW PROBLEMS
R4-5. Determine the x,y, z components of reaction at the
fixed wall A. The 150-N force is parallel to the z axis and the
200-N force is parallel to they axis.
R4-7. The uniform 20-lb ladder rests on the rough floor for
which the coefficient of static friction isµ., = 0.4 and against
the smooth wall at 8. Determine the horizontal force P the
man must exert on the ladder in order to cause it to move.
B
150 N
x
y
8 ft
p
2 Ill
A
.... "..:.... ·.:::. ·.::~
... "..:.-...·.. :
.
.
200 N
...
r.- - - 6 r t
Prob. R4-5
P rob. R4-7
A vertical force of 80 lb acts on the crankshaft.
Determine the horizontal equilibrium force P that must be
applied to the handle and the x, y, z components of reaction
at the journal bearing A and thrust bearing B. The bearings
are properly aligned and exert only force reactions on
the shaft.
*R4-8. The uniform 60-kg crate C rests uniformly on a
10-kg dolly D. If the front wheels of the dolly at A are
locked to prevent rolling while the wheels at B are free to
roll, determine the maximum force P that may be applied
without causing motion of the crate. The coefficient of static
friction between the wheels and the floor is µ.1 = 0.35 and
between the dolly and the crate, µ. 4 = 0.5.
R~.
t-0.6m-I
P-.,...-..-1
c
0.8 m
Prob. R~
Prob. R4-8
1.5 m
CHAPTER
__
6
_,_
(©Tim Scrivener/Alamy)
In order to design the many parts of this boom assembly it is required that
we know the forces that they must support. In this chapter we will show how
to analyze such structures using the equations of equilibrium.
STRUCTURAL
ANALYSIS
CHAPTER OBJECTIVES
•
To show how to determine the forces in the members of a truss
using the method of joints and the method of sections.
•
To analyze the forces acting on the members of a frame or
machine composed of pin-connected members.
5.1
SIMPLE TRUSSES
A truss is a structure composed of slender members joined together at
their end points. The members commonly used in construction consist of
wooden struts or metal bars. In particular, planar trusses lie in a single
plane and are often used to support roofs and bridges. The truss shown in
Fig. 5- la is an example of a typical roof-supporting truss. Here, the roof
load is transmitted to the truss at the joints by means of a series of purlins.
Since this loading acts in the same plane as the truss, Fig. 5- lb, the analysis
of the forces developed in the truss members will be two-dimensional.
Roof truss
(b)
Fig. 5-1
223
224
C H APT ER
5
S TRUCTURAL A NA LYSIS
Floor beam
(a)
t
f
Bridge truss
(b)
Fig. 5-2
\
Gusset
plate
In t he case of a bridge, such as shown in Fig. 5- 2a, the load on t he deck
is first transmitted to stringers, t hen to floor beams, and finally to the
joints of the two supporting side trusses. Like t he roof truss, the bridge
truss loading is also coplanar, Fig. 5- 2b.
When bridge or roof trusses extend over large distances, a rocker or roller
is commonly used for supporting one end, for example, joint A in Figs. 5- la
and 5-2a.This type of support allows freedom for expansion or contraction
of the members due to a change in temperature or application of loads.
\"}•~4':
~ "
Assumptions for Design. To design both the members and the
(a)
connections of a truss, it is first necessary to determine the force
developed in each member when the truss is subjected to a given loading.
To do a force analysis we will make two important assumptions:
•
•
(b)
Fig. 5- 3
A ll loadings are applied at the joints. In most situations, such as for
bridge and roof trusses, this assumption is true. Frequently the weight of
the members is neglected !because the force supported by each member
is usually much larger than its weight. However, if the weight is to be
included in the analysis, it is generally satisfactory to apply it as a vertical
force, with half of its magnitude applied at each end of the member.
The members are joined together by smooth pins. The joint
connections are usually formed by bolting or welding the ends of
the members to a common plate, called a gusset plate, Fig. 5- 3a, or
by simply passing a large bolt or pin through each of t he members,
Fig. 5- 3b. We can assume these connections act as pins provided the
centerlines of the joining members are concurrent, as in Fig. 5- 3.
5.1
T
c
T
c
Tension
(a)
SIM PLE TRUSSES
225
Compression
(b)
Fig. 5-4
Because of these two assumptions, each truss member will act as a twoforce member, and the refore the force acting at each end of the member
will be directed along the axis of the member. If the force tends to
elongate the me mbe r, it is a tensile force (T), Fig. 5-4a; whereas if it tends
to shorten the me mbe r, it is a compressive force (C), Fig. 5-4b. In the
actual design of a truss it is important to state whether the force is tensile
or compressive. Ofte n, compression members must be made thicker than
tension members because of the buckling or sudden collapse that can
occur when a me mbe r is in compression.
Simple Trus
If three membe rs are pin connected at their e nds, they
form a triangular truss that will be rigid, Fig. 5- 5. Attaching two more
members and connecting the m to a new joint D forms a large r truss,
Fig. 5-6. This procedure can be repeated as many times as desired to
form an even larger truss, and by doing this one forms a simple truss.
p
p
Fig. 5-5
c
Fig. 5-6
The use of metal gusset plates in the
construction of these Warren trusses is
clearly evident.
226
CHAPTER
5
STRUCTURAL ANALYSIS
5. 2
2m
- - -2m - - -
(a)
~SOON
FBA (tension)
.
t45~FBc (compression)
(b)
SOON
B
FB C (compression)
FBA (tension)
One way to determine the force in each member of a truss is to use the
method ofjoints. This method is based on the fact that if the entire truss
is in equilibrium, then each of its joints is also in equilibrium. Therefore,
if the free-body diagram of each joint is drawn, the force equilibrium
equations, lFr = 0 and lFy = 0, can then be used to obtain the member
forces acting on each joint.
For example, consider the pin at joint B of the truss in Fig. 5- 7a.
As shown on its free-body dliagram, Fig. 5- 7b, three forces act on the
pin, namely, the 500-N force and the forces exerted by members BA
and BC. Here, FaA is "pulling" on the pin, which means that member
BA is in tension; whereas Fae is " pushing" on the pin, and consequently
member BC is in compression. These effects can also be seen by
isolating the joint with small segments of the member connected to
the pin, Fig. 5- 7c. The pushing or pulling on these small segments
indicates the effect on the members being either in compression
or tension.
When using the method of joints, always start at a joint having at least
one known force and at most two unknown forces, as in Fig. 5- ?b. In this
way, application of lFr = 0 and lFy = 0 yields two algebraic equations
which can be solved for the two unknowns. When applying these
equations, the correct sense of an unknown member force can be
determined using one of two possible methods.
•
The correct sense of direction of an unknown member force
can, in many cases, be determined "by inspection." For example,
Fae in Fig. 5- ?b must iPUSh on the pin (compression) since its
horizontal component, Fae sin 45°, must balance the 500-N force
(lF.r = 0). Likewise, F8 A is a tensile force since it balances the
vertical component, Fae cos 45° (lFy = 0). In more complicated
cases, the sense of an unknown member force can be assumed;
then, after applying the equilibrium equations, the assumed sense
can be verified from the numerical results. A positive scalar
indicates that the sense is correct, whereas a negative scalar
indicates that the sense shown on the free-body diagram must
be reversed.
•
Always assume the unknown member forces acting on the joint's
free-body diagram to be in tension; i.e., the forces "pull" on the pin.
If this is done, then numerical solution of the equilibrium equations
will yield positive scalars for members in tension and negative scalars
for members in compression. Once an unknown member force is
found , use its correct magnitude and sense (T or C) on subsequent
joint free-body diagrams.
(c)
Fig. 5-7
The forces in the members of this simple
roof truss can be determined using the
method of joints.
THE METHOD OF JOINTS
5.2
IMPORTANT POINTS
• Simple trusses are composed of triangular elements. The
members are assumed to be pin connected at their ends and
the loads applied at the joints.
• If a truss is in equilibrium, then each of its joints is in equilibrium.
The internal forces in the members become external forces when
the free-body diagram of each joint of the truss is drawn. A force
pulling on a joint is caused by tension in a member, and a force
pushing on a joint is caused by compression.
PROCEDURE FOR ANALYSIS
The following procedure provides a means for analyzing a truss using
the method of joints.
• Draw the free-body diagram of a joint having at least one
known force and at most two unknown forces. If this joint is at
one of the supports, then it may be necessary first to calculate
the external reactions at the support.
• Orient the x and y axes such that the forces on the free-body
diagram can be easily resolved into their x and y components,
and then apply the two force equilibrium equations 2F, = 0
and 2£,, = 0. Solve for the two unknown member forces and
verify their correct sense.
• Using the calculated results, continue to analyze each of the
other joints. Remember that a member in compression "pushes"
on the joint and a member in tension "pulls" on the joint.
THE METHOD OF JOINTS
227
228
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.1
Determine the force in each member of the truss shown in Fig. 5--8a and
indicate whether the members are in tension or compression.
SOLUTION
We will begin our analysis at joint B since there are one known force
and two unknown forces there.
l -2m-Joint S.
The free-body diagram is shown in Fig. 5- 8b. Applying the
equations of equilibrium, we have
(a)
SOON
B
FsAI~:sc
~ 2F., = O;
+ iIFy
=
O;
500 N - Fae sin 45° = 0
Fae cos45° - FaA = 0
Fae = 707.1 N (C) Ans.
FaA = 500 N (T) Ans.
(b)
Now that the force in member BC has been calculated, we can proceed
to analyze joint C to determine the force in member CA and the support
reaction at the rocker.
Joint C.
(c)
~2F, = O;
FsA =SOON
A
A,-~'6--• FcA
From the free-body diagram, Fig. 5--8c, we have
+ jIF.y
=
-FeA + 707.1cos45° N = 0 FeA = 500 N (T) Ans.
O·, Cy - 707.1 sin 45° N = 0
Ans.
Cy = 500 N
= 500 N
Although it is not necessary, we can determine the
components of the support reactions at joint A using the results of FCA
and FaA- From the free-body diagram, Fig. 5- 8d, we have
Joint A.
(d)
~2F, = O;
SOON
B
+ jIF.y
500N~07.1 N
O·, 500 N - A y
=
0
Ax = 500N
A y = 500 N
NOTE: The results of this analysis are summarized in Fig. 5--8e. Here
4S0
the free-body diagram of each joint (or pin) shows the effects of all
the connected members and external forces applied to the joint,
whereas the free-body diagram of each member shows only the effects
of the joints on the member.
c
0
't;;
c
~
500=i~
soo N
=
500N-A.r = 0
707.lN
Tension 4s• ~
soo....Nl-a::====:::::i~soo N• ~ N
00
SOON
(e)
Fig. 5-8
5.2
I
EXAMPLE
229
THE METHOD OF JOINTS
5.2
Determine the forces acting in all the members of the truss shown
in Fig. 5- 9a, and indicate whether the members are in tension or
compress10n.
SOLUTION
By inspection, there are more than two unknowns at each joint. As a
result, the support reactions on the truss must first be determined.
Show that they have been correctly calculated on the free-body
diagram in Fig. 5- 9b. We can now begin the analysis at joint C.
______
B
~---l~ 3
kN
1
2m
LA.
-
2m - -2m -
(a)
Joint C.
From the free-body diagram, Fig. 5- 9c,
~2F.x = O·,
-FcD cos 30° + FcB sin 45° = 0
+ j2F.y
1.5 kN + FcD sin 30° - FcB cos 45°
=
O·,
=
0
These two equations must be solved simultaneously for each of the two
unknowns. A more direct solution can be obtained by applying a force
summation along an axis that is perpendicular to the direction of the
other unknown force. For example, summing forces along the y' axis,
which is perpendicular to the direction of FcD, Fig. 5- 9d, yields a direct
solution for FcB·
1.5 cos 30° kN - FcB sin 15° = 0
FcB = 5.019 kN = 5.02 kN (C)
3kN
-
2m - -2m -
1.5 kN
1.5 kN
(b}
y
Fcs
Ans.
45•
1.5 kN
(c)
Then,
Fcs
-FcD + 5.019 cos 15° - 1.5 sin 30° = O;
FcD = 4.10 kN (T) Ans.
x'
30°
Joint 0.
We can now proceed to analyze joint D. The free-body
diagram is shown in Fig. 5- 9e.
~2F.x = O·,
+ j2F.y
=
O·,
-FDA
30° + 4.10 cos 30° kN
FDA = 4.10 kN (T)
COS
=
1.5 kN
(d)
y
0
I Fos
Ans.
FDB - 2(4.10 sin 30° kN) = O
FDB = 4.10 kN
(T)
Ans.
NOTE: The force in the last member, BA, can be obtained from joint B or
joint A. As an exercise, draw the free-body diagram of joint B, sum the
forces in the horizontal direction, and show that FBA = 0.776 kN (C).
FoA
4.lOkN
(e)
Fig. 5- 9
230
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.3
Determine the force in each member of the truss shown in Fig. 5- lOa.
Indicate whether the members are in tension or compression.
400N
400N
c
C1
3mC
4m
D
600N
1 - - - -6 m - - - - 1
- 3m-1--3m-
600N
(b)
(a)
Fig. 5-10
SOLUTION
Support Reactions. No joint can be analyzed until the support
reactions are determined, because each joint has at least three
unknown forces acting on it. A free-body diagram of the entire truss is
given in Fig. 5- lOb. Applying the equations of equilibrium, we have
~IF.x = O·,
C+IMc
O;
The analysis can now start at either joint A or C. The choice is arbitrary
since there are one known and two unknown member forces acting on
the pin at each of these joints.
y
1
=
600 N - Cx = 0
Cx = 600 N
-Ay(6 m) + 400 N(3 m) + 600 N(4 m) = 0
A y = 600N
600 N - 400 N - Cy = 0
Cy = 200 N
FAB
4
F AD
A<:""-----1~
600N
- - -x
Joint A. (Fig. 5- lOc). As shown on the free-body diagram, FAB is
assumed to be compressive and F AD is tensile. Applying the equations
of equilibrium, we have
+f IFy
(c)
=
~IF.x =
0;
O·,
600N - ~FAB = 0
FAD - ~(750 N) = 0
FAB =
FAD =
750 N (C)
450 N (T)
Ans.
Ans.
5.2
THE METHOD OF JOINTS
(Fig. 5- lOd). Using the result for FAD and summing forces in
the horizontal direction, we have
Joint 0.
..:!;. IF,
-450N + ~FDB + 600N
O;
=
=
0
FDB = -250N
The negative sign indicates that F DB acts in the opposite sense
shown in Fig. 5- lOd. * Hence,
FDB = 250 N (T)
~o
that
450N D
600N
(d)
Ans.
To determine F De, we can either correct the sense of F DB on the freebody diagram, and then apply 'I.Py = 0, or apply this equation and
retain the negative sign for FDB• i.e.,
-FDc - ~(-250N) = 0
+ fl£,, = O;
Joint C.
FDc = 200N
(C)
1
Ans.
1200N
(Fig. 5- lOe).
Fcs
..:!;. IF.x = O·,
+ fl£,, = 0;
FcB - 600 N = 0
200 N - 200 N
FcB = 600 N
=0
(C)
Ans.
(check)
body diagram for each joint and member.
400N
600 N Compression
250N
750N
6
A
250 N ' !200 N
Tension
-+-a::========:: D- - •
~N
c
~ON
D
f
600N _ x
200N
NOTE: The analysis is summarized in Fig. 5- lOf, which shows the free-
750N
~~ c
• 600 N
600N
(f)
Fig. 5-10 (cont.)
*The proper sense could have been determined by inspection, prior to applying "i,F, = 0.
I
(e)
2 31
232
CHAPTER
5
STRUCTURAL ANALYSIS
5.3 ZERO-FORCE MEMBERS
Truss analysis using the method of joints is greatly simplified if we can
first identify those members which support no loading. These zero-force
members are used to increase the stability of the truss during construction
and to provide added support if the loading is changed.
The zero-force members of a truss can generally be found by
inspection of each of the joints. For example, consider the truss shown
in Fig. 5- lla. If a free-body diagram of the pin at joint A is drawn,
Fig. 5- 1 lb, it is seen that members AB and AF are zero-force members.
(We could not come to this conclusion if we had considered the
free-body diagrams of joints F or B, simply because there are five
unknowns at each of these joints.) In a similar manner, consider the
free-body diagram of joint D , Fig. 5- llc. Here again it is seen that DC
and DE are zero-force members. To summarize, then, if only two noncollinear members form a truss joint and no external load or support
reaction is applied to the joint, the two members must be zero-force
members. The load on the truss in Fig. 5- lla is therefore actually
supported by only five members, as shown in Fig. 5- 1 ld.
D
1
A
c
A
L
- -x
FAB
+ lF, = O; FAB = 0
+i 2Fy = O; FAF= 0
B
~
p
(a)
(b)
D
+ \, 2Fy = O; FDC sin 0 = O; FDC = 0 since sin 9
+iClF, = 0; FDE + 0 = 0; FDE = 0
F
*0
E
B
p
(d)
(c)
Fig. 5-11
5.3
ZERO-FORCE M EMBERS
Now consider the truss shown in Fig. 5-12a. The free-body diagram of
the pin at joint D is shown in Fig. 5-12b. By orienting they axis along
members DC and DE and the x axis along member DA, it is seen that DA
is a zero-force member. From the free-body diagram of joint C, Fig. 5-12c,
it can be seen that this is also the case for member CA. In general then, if
three members form a truss joint for which two ofthe members are collinear,
the third member is a zero-force member provided no external force or
support reaction has a component that acts along this member. The truss
shown in Fig. 5-12" is therefore suitable for supporting the load P.
£
p
Foe
/
FoA
"""
y
x
+ .C 'S.F.., = 0;
+'>i '2Fy = O;
B
(a)
FoA • 0
F0 e = FOi-:.
(b)
p
£
Fes
y
x
+.c 'f.Fx =
0:
+'>i 'f.Fy = 0:
Fe,.. sin 0 = 0:
Fes = Feo
A
FeA = 0 since sin 0 '# O;
(c)
(d)
Fig.5-U
IMPORTANT POINT
• Zero-force members support no load; however, they are
necessary for stability, and are available when additional
loadings are applied to the joints of the truss. These members
can usually be identified by inspection. They occur at joints
where only two mernbers are connected and no external load
acts along either member. Also, at joints having two collinear
members, a third member will be a zero-force member if no
external force components act along this member.
233
234
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.4
Determine all the zero-force members of the Fink roof truss shown in
Fig. 5- 13a. Assume all joints a.re pin connected.
SkN
x
(b}
c
x
(a)
(c)
Fig. 5-13
SOLUTION
Look for joint geometries that have three members for which two are
collinear. We have
x
(d}
Joint G.
(Fig. 5- 13b).
+ jIF.y
O·,
=
Ans.
Fee = 0
Realize that we could not conclude that GC is a zero-force member by
considering joint C, where there are five unknowns. The fact that GC
is a zero-force member means that the 5-kN load at C must be
supported by members CB, CH, CF, and CD.
\
(Fig. 5- 13c).
+il'IF,
O;
=
Ans.
FvF = 0
x
(e)
Joint F.
+ jIF.y
y
(Fig. 5- 13d).
=
O·,
FFc cos 8 = 0
Since 8 # 90°,
FFc = 0
Ans.
NOTE: If joint B is analyzed, Fig. 5- 13e,
2kN
FHc
FHA
Joint 0 .
H
(f }
-
FHG
+ \.IF...t = O·,
2kN -
FBH =
0
FBH =
2 kN (C)
x
Also, FHc must satisfy IF). = 0, Fig. 5-13/, and therefore HC is not a
zero-force member.
5.3
2 35
ZERO-FORCE MEMBERS
PRELIMINARY PROBLEMS
P5-L In each case, calculate the support reactions and
then draw the free-body diagrams of joints A , B , and C of
the truss.
E
P5-2. Identify the zero-force members in each truss.
D
•
H
G
F
E
2m
A
~=======3t.======~•
8
- - -2
--'-
A
m---;1-- - -2 m- lc
400N
(a)
(a)
E
E
4m
A "
,__ _ 2m - - -- •- - - - 2m - - -
"
1c
IB
l - 2m-----l-2m-l-2m SOON
600N
(b)
Prob. P5-l
D
(b)
Prob. P5-2
700N
236
C H APT ER
5
S TRUCTURAL A NA LYS IS
FUNDAMENTAL PROBLEMS
All problem solutions must include FBDs.
Determine the force in each member of the truss
and state if the members are in tension or compression.
F5-L
FS-4. Determine the greatest load P that can be applied
to the truss so that none of the members are subjected to a
force exceeding either 2 kN in tension or 1.5 kN in
compression .
- - 4 ft - - 1; - -4ft
p
450 Jb
Prob.FS-1
FS-2. Determine the force in each member of the truss
and state if the members are in tension or compression.
- -3m - - -
Prob.FS-4
FS-5. Identify the zero-force members in the truss.
3 kN
- - - i - - -2 m - - -1
D
3 ft
- [ D/o~=:::::A===;::j>c
1.5 nn
- -2
l_ IBI;======~
ft--1~ 2 ft
A
300 Jb
Prob.FS-5
Prob.FS-2
FS-3. Determine the force in each member of the truss
and state if the members are in tension or compression.
FS-6. Determine the force in each member of the truss
and state if the members are in tension or compression.
c
D
·1
Q
6001b
450 lb
E
3 ft
J
A
A
4 ft
800 Jb
6001b
Prob.FS-3
oc
0
- -3
ft--1~3 ft - Prob.FS-6
5.3
237
ZERO-FORCE MEMBERS
PROBLEMS
All problem solutions must include FBDs.
5-1. Determine the force in each member of the truss and
state if the members are in tension or compression. Set
P 1 =20 kN, P2 =10 kN.
5- 2. Determine the force in each member of the truss and
state if the members are in tension or compression. Set
P 1 =45 kN,P2 =30kN.
*5-4. Determine the force in each member of the truss
and state if the members are in tension or compression.
3
kip
-1- 2 kip
8 ft
i
4 ft
H
3 kip
i.s kip~f'OK
-1
0£
ID ~10ft --10ft --10ft ~10ft~
A-
c
B
Prob.5-4
5-5. Determine the force in each member of the truss, and
state if the members are in tension or compression. Set 9 = 0°.
5-6. Determine the force in each member of the truss, and
state if the members are in tension or compression. Set 9 = 30°.
D
-.-1------7<.!~-.-- 3kN
i - - - - - - - 2 m _ _ __,
Probs. 5-112
1.5 m
_]A•
B
1 - - - 2 m - - - •-- -2 m - - - i
4 kN
Probs. 5-5/6
5- 3. Determine the force in each member of the truss and
state if the members are in tension or compression.
5-7. Determine the force in each member of the truss and
state if the members are in tension or compression.
4kN
- 3m-
B
3mc l 3m-I
Sm
SkN
Prob.5- 3
Prob. 5-7
238
CHAPTER
5
S TRUCTURAL A NA LYS IS
*5-8. Determine the force in each member of the truss in
terms of the load P and state if the members are in tension
or compression.
5-13. Determine the force in each member of the truss
and state if the members are in tension or compression. Set
P 1 =10kN, P 2 =8kN.
5-9. Members AB and BC can each support a maximum
compressive force of 800 lb, and members AD,DC, and BD
can support a maximum tensile force of 1500 lb. If a= 10 ft,
determine the greatest load P the truss can support.
5-14. Determine the force in each member of the truss
and state if the members are in tension or compression. Set
P 1 =8 kN,P2 =12kN.
5-10. Members AB and BC can each support a maximum
compressive force of 800 lb, and members AD,DC, and BD
can support a maximum tensile force of 2000 lb. If a= 6 ft,
determine the greatest load P the truss can support.
G
F
E
T
2m
B
Probs. 5-13/14
p
5-15. Determine the force in each member of the truss
and state if the members are in tension or compression. Set
P 1 =9 kN,P2 =15 kN.
Probs. 5-8/9/10
5-11. Determine the force in each member of the truss
and state if the members are in tension or compression. Set
P=8kN.
*5-12. If the maximum force that any member can support
is 8 kN in tension and 6 kN in compression, determine the
maximum force P that can be supported at joint D.
*5-16. Determine the force in each member of the truss
and state if the members are in tension or compression. Set
P 1 =30kN, P 2 =15 kN.
E
D
4m
1 - - - 3 m - - - 1-- - 3 m -----..1
- - 4m - - - - 4m - - o
p
Probs. 5-11/12
Probs. 5-15/16
5.4
5.4
THE METHOD OF SECTIONS
When we need to find the force in only a few members of a truss, we can
analyze the truss using the method of sections. It is based on the principle
that if the truss is in equilibrium then any part of the truss is also in
equilibrium. For example, consider the two truss members shown in
Fig. 5-14. If the forces within the members are to be determined, the n an
imaginary section, indicated by the blue line, can be used to cut each member
into two parts and thereby "expose" each internal force as "external" to the
free-body diagrams of the parts shown on the right. Clearly, it can be seen
that equilibrium requires that the member in tension (T) be subjected to a
"pu!J,'' whereas the member in compression (C) is subjected to a "push."
The me thod of sections can also be used to "cut" or section the members
of an entire truss. If the section passes through the truss and the free-body
diagram of eithe r of its two parts is drawn, we can then apply the equations
of equilibrium to that part to determine the member forces at the section.
Since only three independe nt equilibrium equations (IF.r = 0, I.Py = 0,
2.M0 = 0) can be applied to the free-body diagram of any part, then we
should try to select a section that, in general, passes through not more
than three me mbers in which the forces are unknown. For example,
consider the truss in Fig. 5- 15a. If the forces in members BC, GC, and GF
are to be determined, then section aa would be appropriate. The freebody diagrams of the two parts are shown in Figs. 5-15b and 5- 15c. Notice
that the direction of each member force is specified from the geonietry of
the truss, since the force in a member is along its axis. Also, the membe r
forces acting on one part of the truss are equal but opposite to those
acting on the other part - Newton's third law. Members BC and GC are
assumed to be in tension since they are subjected to a "pull,'' whereas GF
is in compression since it is subjected to a "push."
The three unknown member forces F80 Fee, and FcFcan be obtained
by applying the three equilibrium equations to the free-body diagram in
Fig. 5-15b. Ir, however, the free-body diagram in Fig. 5-15c is conside red,
then the three support reactions Dx, D,, and Ex will have to be known,
because only three equations of equilibrium are available. (This, of
course, is done in the usual manner by considering a free-body diagram
of the entire truss.)
a
c
T
T
T
lntemal
tensile
forces
Tension
T
c
c
c
Lnterna
compressive
forces
Zm-
0
l- 2 m-
IOOON
F
c
Compression
Dy
t
2m
I
Fae
G /---=--!==::la=:===lll-~L- E,
E
Fap
-
2m -
lOOON
(a)
c
Fig. 5-14
. -I
l- 2 m-I
c
c 1- 2 m- - ,q.........,._.
Af==.=~==l=~=~l -l
G
T
T
2 111
-
239
TH E M ETHOD OF SECTIONS
(b)
Fig. 5-15
(c)
240
CHAPTER
5
S TRUCTURAL A NA LYSIS
- 2m-1 C
l
Fsc
----.1---,.p;::~+.7
2m
_l
¢::::::::==£=:'.::I+-::.
- 2m-lG
lOOON
(b)
Fig. 5-15 (Repeated)
When applying the equilibrium equations, we should carefully
consider ways of writing the equations so as to yield a direct solution for
each of the unknowns, rather than having to solve simultaneous
equations. For example, using the part in Fig. 5- 15b and summing
moments about C will yield a direct solution for FeF since F8 c and Fee
create zero moment about C. Likewise, F8 c can be directly obtained by
summing moments about G. Finally, Fee can be found directly from a
force summation in the vertical direction since FeF and F 8 c have no
vertical components. This ability to determine directly the force in a
particular truss member is one of the main advantages of using the
method of sections.*
As in the method of joints, there are two ways in which we can
determine the correct sense of an unknown member force:
•
The correct sense can in many cases be determined "by inspection."
For example, FBC is shown as a tensile force in Fig. 5- 15b since
moment equilibrium about G requires that F8 c create a moment
opposite to that of the 1000-N force. Also, Fee is tensile since its
vertical component must balance the 1000-N force which acts
downward. In more complicated cases, the sense of an unknown force
may be assumed. If the solution yields a negative scalar, it indicates
that the force's sense of direction is opposite to that shown on the
free-body diagram.
•
Always assume that the unknown member forces at the section are
tensile forces, i.e., "pulling" on the member. By doing this, the
numerical solution of the equilibrium equations will yield positive
scalars for members in tension and negative scalars fo r members in
compression.
*If the method of joints were used t o determine, say, the force in member GC, it would
be necessary to analyze joints A, B, and G in sequence.
The forces in selected members of this
Pratt truss ,can readily be dete rmined
using the method of sections.
5.4
TH E M ETHOD OF SECTIONS
IMPORTANT POINT
• If a truss is in equilibrium, then each of its parts is in equilibrium.
The internal forces in the members become external forces
when the free-body diagram of a part of the truss is drawn. A
force pulling on a member causes tension in the member, and a
force pushing on a member causes compression.
Simple trusses are often used in
the construction of large cranes
in order to reduce the weight of
the boom and tower.
PROCEDURE FOR ANALYSIS
The forces in the members of a truss may be determined by the
method of sections using the following procedure.
Free-Body Diagram.
• Make a decision as to how to "cut" or section the truss through
the members where forces are to be determined.
• Before isolating any part of the truss, it may first be necessary to
detennine the truss's support reactions that act on the part. Once
this is done then the three equilibrium equations will be available
to solve for member forces at the section.
• Draw the free-body diagram of that part of the sectioned truss
which has the least number of forces acting on it.
Equations of Equilibrium.
• Moments should be summed about a point that lies at the
intersection of the lines of action of two unknown forces, so that
the third unknown force can be determined directly from the
moment equation.
• If two of the unknown forces are parallel, forces may be summed
perpendicular to their direction in order to directly determine the
third unknown force.
241
242
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.5
Determine the force in members GE, GC, and BC of the truss shown in
Fig. 5-16a. Indicate whether the members are in tension or compression.
E
a
SOLUTION
Section aa in Fig. 5- 16a has been chosen since it cuts through the three
members whose forces are to be determined. In order to use the
method of sections, however, it is first necessary to determine
the external reactions at A or D. Why? A free-body diagram of
the entire truss is shown in Fig. 5- 16b. Applying the equations of
equilibrium, we have
1200N
(a)
~2F, = O;
400N-Ax = 0
Ax = 400N
-1200 N(8 m) - 400 N(3 m) + Dy(12 m) = 0
I
Dy = 900N
3m
~JA~===lt=::==\t==~~D
A.r
1 - - -8 01
Ay
-
4m
1200N
(b)
+ j2F.y
Ay - 1200 N + 900 N = 0
O·,
=
Ay = 300N
For the analysis the free-body diagram of the
left part of the sectioned truss will be used, since it involves the least
number of forces, Fig. 5- 16c.
Free-Body Diagram.
Equations of Equilibrium. Summing moments about point G
eliminates Fee and Fee and yields a direct solution for Fae·
C + 2Me = O;
-300N(4m) - 400N(3m) + Fae (3m) = 0
Fae = 800 N
FGc
3m
IA
' c
400N
Fsc1
-
4 m - - 4m - I
C+ 2Me =
O;
-300 N(8 m) + Fee(3 m) = 0
Fee = 800 N
Fig. 5-16
Ans.
In the same manner, summing moments about point C we obtain a
direct solution for Fee·
300N
(c)
(T)
(C)
Ans.
Since Fae and Fee have no vertical components, summing forces in
they direction directly yields Fee, i.e.,
+ j2F.y
=
O·,
300N - ~Fee = 0
Fee = 500 N
(T)
Ans.
NOTE: Here it is possible to tell, by inspection, the proper direction for
each unknown member force. For example, 2Me = 0 requires Fee to
be compressive because it must balance the moment of the 300-N
force about C.
5.4
I
EXAMPLE
243
THE METHOD OF SECTIONS
5.6
Determine the force in member CF of the truss shown in Fig. 5- 17a.
Indicate whether the member is in tension or compression. Assume
each member is pin connected.
G
B
l-4 m
a
4 m - t - 4 m - - 4m - I
5kN
3 kN
1 - - -8 m - - -·-•.:___ 4 m
3.25 kN
5kN
(a)
___::!~ 4 m .:___
3kN
4.75kN
(b)
Fig. 5-17
SOLUTION
Section aa in Fig. 5- 17a will be used since this
section will "expose" the internal force in member CF as "external" on
the free-body diagram of either the right or left portion of the truss. It
is first necessary, however, to determine the support reactions on either
the left or right side. Verify the results shown on the free-body diagram
in Fig. 5- 17b.
Free-Body Diagram.
The free-body diagram of the right part of the truss, which is the easiest
to analyze, is shown in Fig. 5- 17c. There are three unknowns,
FFG• FcF, and FcD·
G
~,----------1~
Equations of Equilibrium. We will apply the moment equation
about point 0 in order to eliminate the two unknowns FFc and
FcD· The location of point 0, measured from E, can be determined
from proportionaltriangles,i.e.,4/ (4 + x) = 6/ (8 + x), x = 4 m.
Or, stated in another manner, since the slope of member GF has a
drop of 2 m to a horizontal distance of 4 m, and FD is 4 m,
Fig. 5- 17c, then from D to 0 the distance must be 8 m.
I
+ (3kN)(8m) - (4.75kN)(4m)
FcF = 0.589 kN
(C)
I
6m: FCF
FCF cos45°C~
4m
.
· ~.=a~' ',',,o_l
45~t- 4 ~~·D_TD 4 m E x
FcFsin 45° f
3 kN
4.75 kN
(c)
An easy way to determine the moment of FcF about point 0 is to
use the principle of transmissibility and slide FcF to point C, and
then resolve FcF into its two rectangular components. We have
C+lM0 = O;
-FcFsin45°(12m)
2m
= 0
Ans.
-=i
244
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.7
Determine the force in member EB of the roof truss shown in Fig. 5- 18a.
Indicate whether the member is in tension or compression.
1000 N
3000 N
lOOON
b
Eb
SOLUTION
lOOON
By the method of sections, any imaginary section
that cuts through EB will also have to cut through three other members
for which the forces are unknown. For example, section aa cuts through
ED, EB, FB,and AB. If a free-body diagram of the left part of this section
is considered, Fig. 5- 18b, it is poosible to obtain FED by summing moments
about B to eliminate the other three unknowns; however, FEB cannot be
determined from the remaining two equilibrium equations.
One possible way of obtaining FEB is first to determine FED from
section aa, then use this result on section bb, Fig. 5- 18a, which is shown in
Fig. 5- 1&. Here the force system is concurrent and our sectioned freebody diagram is the same as the free-body diagram for the joint at E.
Free-Body Diagrams.
c
A
4::'.:::::::::::::::::::)::::::::=ll::::::::====~
-
2m - -2m - -2m - -2m 2000N
4000N
(a)
lOOON
3000N
y
E
I
lOOON
lOOON
FEo = 3000N
(b)
Fig.5-18
In order to determine the moment of
FED about point B, Fig. 5- 18b, we will use the principle of transmissibility
and slide this force to point C and then resolve it into its rectangular
components as shown. Therefore,
C +lMB = O;
1000N(4m) + 3000N(2m) - 4000N(4m)
Equations of Equilibrium.
+ FED sin 30°(4 m)
FED
=
O
3000 N (C)
Considering now the free-body diagram of section bb, Fig. 5- 1&, we have
~ IF, = O;
FEF cos 30° - 3000 cos 30° N
FEF
+ jlFy
=
O;
=
=
3000 N
=
0
(C)
2(3000sin 30°N) - lOOON - FEB = 0
FEB
=
2000 N
(T)
Ans.
5.4
245
THE METHOD OF SECTIONS
FUNDAMENTAL PROBLEMS
All p roblem solutions must include FBDs.
FS-7. Determine the force in members BC, CF, and FE
and state if the members are in tension or compression.
G
F
E
l[jx:·=~~·
FS-10. Determine the force in members EF, CF, and BC
of the truss and state if the members are in tension or
compression.
F
--1
4
ft
B
IDr-;::::=4=ft==i¢s~=4=ft~=-~-c~=:_=4=f=t:_~~Dl
600 lb
600 lb
- 6tt- - 6ft
300lb
300lb
Prob.FS-10
800 lb
Prob. FS-7
FS-11. Determine the force in members GF, GD, and CD
of the truss and state if the members are in tension or
compression.
FS-8. Determine the force in members LK, KC, and CD
of the Pratt truss and state if the members are in tension or
compression.
K
L
J
G
I
l -2 m-
B
C
B
C
- 2 m-
- 2
m-
lOkN
20 kN 30 kN 40 kN
FS-9. Determine the force in members Kl, KD, and CD
of the Pratt truss and state if the members are in tension or
compression.
K
J
I
I
3
01
A
•
B
Prob. FS-11
FS-12. Determine the force in members DC, HI, and JI of
the truss and state if the members are in tension or
compression. Suggesrion: Use the sections shown.
Prob. FS-8
L
C
_Q i--9ft-~6
I
6 ;ft
I
6ft
Ift
ftl F6 tt--j-9ft-1 E
I
s ~++-~
Prob. FS-9
s
I ---1+..'5"----1-:-~Mll-- I
I
12
1-2 m--2 m--2 m -2 m--2 m- -2 m-
20 kN 30 kN 40 kN
m-1
15 kN
25kN
1-2 m--2 m--2 m -2 m- -2 m- -2 m-
D
- 2
l-6ft-l-6tt-1
Prob.FS-12
1600 lb
246
CHAPTER
5
S TRUCTURAL A NA LYS IS
PROBLEMS
A ll problem solutions must include FBDs.
5-17. Determine the force in members DC, HC, and HI of
the truss and state if the members are in tension or
compression.
5-18. Determine the force in members ED, EH, and GH
of the truss and state if the members are in tension or
compression.
5-22. Determine the force in members EI and JI of the
truss which serves to support the deck of a bridge. State if
these members are in tension or compression.
50kN
-2m--2m--2m
40kN
5-21. Determine the force in members CD, CJ, Kl, and
DJ of the truss which serves to support the deck of a bridge.
State if these members are in tension or compression.
4000lb
B~
A
5000lb
8000 lb
C
D
E
~~~~~~:::::::;~~G-r
1
12 ft
1.5 m
C
30kN
J
1
1.5 m
B
J
~~~:::::::::::::~~--1-l
- 9 ft-IL 9 t1-l~9 ft -1! 9 t1 -l~ 9 ft-1~9 ft-
40 kN
Probs. 5-21122
1
l.5m
J
5-23. The Howe 1russ is subjected to the loading shown.
Determine the force in members GF, CD, and GC and
state if the members are in tension or compression.
Probs. 5-17n8
5-19. Determine the force in members HG, HE, and DE
of the truss and state if the members are in tension or
compression.
*5-24. The Howe 1russ is subjected to the loading shown.
Determine the force in members GH, BC, and BG of the
truss and state if the members are in tension or compression.
*5-20. Determine the force in members CD , HI, and CH
of the truss and state if the members are in tension or
compression.
5kN
G
-,r----S'k
~
·N
u---/~~
5 kN
3m
J:t==~~f¢=.~it=~l;;=::::::PF
1-3
ft
- 3ft - -3 ft - - 3ft
3 ft -
1500 lb 1500 lb 1500 lb 1500 lb 1500 lb
Probs. 5-19/20
tkN
4
~
I 2~mJPs=.,,1_[~2m::[==:2m~I'
Probs. 5-23/24
5.4
5-25. Determine the force in members EF. CF, and BC,
and state if tbc members arc in tension or compression.
5-26. Determine the force in members AF, BF, and BC,
and state if the members arc in tension or compression.
247
THE M ETHOD OF SECTIONS
5-29. Determine tbc force in members BC, HC, and HG.
After the truss is sectioned use a single equation of
equilibrium for the calculation of each force. State if these
members arc in tension or compression.
5-30. Determine the force in members CD, CF, and CG
and state if these members are in tension or compression.
4kN
2m
5kN
4 kN
4 kN
3 kN
2 kN
F
8kN
B
c
J
.) Ill
1
2m
2m
.J
A
- 5m--5mj-5m15 mProbs. 5-29130
Probs. 5-25126
5-27. Determine the force in members EF, BE, BC, and
BF of the truss and state if these members are in tension or
compression. Set P 1 =9 kN, P2 = 12 kN, and P3 = 6 kN.
5-31. Determine the force developed in members FE, EB,
and BC of the truss and state if these members are in
tension or compression.
*5-28. Dctem1ine the force in members BC, BE, and EF
of the truss and state if these members are in tension
or compression. Set P 1 =6 kN, P2 = 9 kN, and P3 = 12 kN.
1 2
m~- 1.5 m E j -- 2 m - - - I
E
F
I
3 Dl
AA~~~~~J_
~
C
B
-3 m --3 m -
-
3 m - -I
1
2m
L
c
B
II kN
Probs. 5-27128
22 kN
Pro b. 5-31
248
CHAPTER
5
STRUCTURAL ANALYSIS
5.5
FRAMES AND MACHINES
Frames and machines are two types of structures which are often
composed of pin-connected multiforce members, i.e., members that are
subjected to more than two forces. Frames are used to support loads,
whereas machines contain moving parts and are designed to transmit and
alter the effect of forces. Provided a frame or machine contains no more
supports or members than are necessary to prevent its collapse, then the
forces acting at the joints and supports can be determined by applying the
equations of equilibrium to each of its members. Once these forces are
obtained, it is then possible to design the size of the members, connections,
and supports using the theory of mechanics of materials and an appropriate
engineering design code.
Free-Body Diagrams. In order to determine the forces acting at
This crane is a typical
example of a framework.
Common tools such as these pliers
act as simp le machines. Here the
applied force on the handle s
cre ates a much larger force at
the jaws.
the joints and supports of a frame or machine, the structure must be
disassembled and the free-body diagrams of its parts must be drawn. The
following important points must be observed:
•
Isolate each part by drawing its outlined shape. Then show all the
forces and/or couple moments that act on the part. Make sure to
label or identify each known and unknown force and couple moment
with reference to an established x, y coordinate system. Also,
indicate any dimensions used for taking moments. As usual, the
sense of an unknown force or couple moment can be assumed.
•
Identify all the two-force members in the structure and represent
their free-body diagrams as having two equal but opposite collinear
forces acting at their points of application. (See Sec. 4.4.) By doing
this, we can avoid solving an unnecessary number of equilibrium
equations.
•
Forces common to any two contacting members act with equal
magnitudes but opposite sense on the free-body diagrams of the
respective members.
The following two examples graphically illustrate how to draw the
free-body diagrams of a dismembered frame or machine. In all cases, the
weight of the members is neglected.
5.5
EXAMPLE
249
FRAMES ANO MACHINES
5 .~
For the frame shown in Fig. 5- l 9a, draw the free-body diagram of (a) each
member, (b) the pins at Band A , and (c) the two members connected
together.
n,
p
8
Effecl of
member BC
the pin
-:;:Jon
x
D
Effecl of
member AB
on the pin
8
B,
81
Pin B
c,.
(a)
(c)
(b)
SOLUTION
By inspection, members BA and BC are not two-force
members. Instead, as shown on the free-body diagrams, Fig. 5-19b, BC
is subjected to a force from each of the pins at B and C and the
external force P. Likewise, AB is subjected to a force from each of the
pins at A and Band the external couple moment M. The pin forces are
represented by their x and y components.
Part (a).
The pin at Bis subjected to only two forces, i.e., the force of
member B C and the force of member AB. For equilibriunz these
forces (or their respective components) must be equal but opposite,
Fig. 5-19c. Notice that Newton's third law is applied between the pin
and its connected members, i.e., the effect of the pin on the two
members, Fig. 5-19b, and the equal but opposite effect of the two
members on the pin, Fig. 5-19c. ln the same manner, there are three
forces on pin A , Fig. 5-19d,caused by the force components of member
AB and each of the two pin leaves.
Part (b).
The free-body diagram of both members connected
together, yet removed from the supporting pins at A and C, is shown
in Fig. 5-19e. The force components Bx and Bv are not shown on this
diagram since they are internal forces (Fig.' 5- 19b) and therefore
cancel out. Also, to be consistent when later applying the equilibrium
equations, the unknown force components at A and C in Fig. 5-19e
must act in the same sense as those shown in Fig. 5- 19b.
Part (c).
Pin A
(d)
f'
*P
l\1
\
~
~
A,--•
""4-- -
.1fio1
c,
(c)
Fig. 5-19
ex
250
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.9
For the frame shown in Fig. 5- ZOa, draw the free-body diagrams of (a) the
entire frame including the pullleys and cords, (b) the frame without the
pulleys and cords, and (c) each of the pulleys.
D
75 lb
(a)
SOLUTION
Part (a). When the entire frame including the pulleys and cords is
considered, the interactions at the points where the pulleys and cords are
connected to the frame become pairs of internal forces which cancel each
other and therefore are not shown on the free-body diagram, Fig. 5- 20b.
When the cords and pulleys are removed, their effect on
the frame must be shown, Fig. 5- 20c.
Part (b).
Part (c). The force components Bx, B>• C-"' C>' of the pins on the
pulleys are equal but opposite to the force components exerted by the
pins on the frame, Fig. 5-20c.
T
A,
f
Ay
By
T
•
'J
B, -.li:====~=::!!::!!::~===~
· ,. _:-
-
-
11
"
75 lb
(b)
(c)
Fig. 5-20
A,
5.5
PROCEDURE FOR ANALYSIS
The joint reactions on frames or machines (structures) composed of
multiforce members can be determined using the following procedure.
Free-Body Diagram.
• Draw the free-body diagram of the entire frame or machine, a
portion of it, or each of its members. The choice should be
made so that it leads to the most direct solution of the problem.
• When the free-body diagram of a group of members of a frame
or machine is drawn, the forces between the connected parts of
this group are internal forces and are not shown on the free-body
diagram of the group.
• Forces common to two members which are in contact act with
equal magnitude but opposite sense on the respective free-body
diagrams of the members.
• A two-force member, regardless of its shape, has equal but
opposite collinear forces acting at the ends of the member.
• In many cases it is possible to tell by inspection the proper sense
of the unknown forces acting on a member; however, if this seems
difficult, the sense can be assumed.
• Remember that a couple moment is a free vector and can act
at any point on the free-body diagram. Also, a force is a sliding
vector and can act at any point along its line of action.
Equations of Equilibrium.
• Count the number of unknowns and compare it to the total
number of equilibrium equations that are available. In two
dimensions, there are three equilibrium equations that can be
written for each member.
• Sum moments about a point that lies at the intersection of the
lines of action of as many of the unknown forces as possible.
• If the solution of a force or couple moment is found to be a
negative scalar, it means the sense of the force is the reverse of
that shown on the free-body diagram.
FRAMES ANO MACHINES
2 51
252
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
5.10
I
Determine the horizontal and vertical components of force which the pin
at C exerts on member BC of the frame in Fig. 5- 21a.
SOLUTION I
Free-Body Diagrams. By inspection it can be seen that AB is a
two-force member. The free-body diagrams are shown in Fig. 5- 21b.
Equations of Equilibrium. The three unknowns can be determined
by applying the three equations of equilibrium to member CB.
C+ lMc = 0; 2000 N(2 m) - (FAB sin 60°)(4 m) = 0 FAB = 1154.7 N
(a)
~IF, = O; 1154.7 cos 60° N - Cx = 0
Cx = 577 N
+ f IF,. = O; 1154.7 sin 60° N - 2000 N +
c,. =
Ans.
0
Cy = 1000 N
Ans.
SOLUTION II
Free-Body Diagrams. If one does not recognize that AB is a
two-force member, then more work is involved in solving this problem.
The free-body diagrams are shown in Fig. 5- 21c.
Equations of Equilibrium. The six unknowns are determined by
applying the three equations of equilibrium to each member.
Member AB
C+IMA
=
O;
~IF, = O;
(b)
+f l£,. = 0; Ay - By = 0
C+lMc
c
/;
2m - l - 2m -
B,
;t
A.T
c,
,
By -
~(c)
Fig. 5-21
=
0
(1)
(2)
(3)
Member BC
2000N
B,
Bx(3 sin 60° m) - B,.(3 cos 60° m)
Ax - Bx = 0
Cy
=
O; 2000N(2m) - By(4m)
~ IF, = O;
Bx - Cx = 0
+ f IF,. = O;
By - 2000 N
=
0
(4)
(5)
+
Cy = 0
(6)
The results for Cx and Cy can be determined by solving these equations
in the following sequence: 4, 1, 5, then 6. The results are
By = lOOON
Bx = 577N
Cx = 577N
Ans.
Ans.
C y = lOOON
By comparison, Solution I is simpler since the requirement that
FAB in Fig. 5-21b be equal, opposite, and collinear at the ends of
member AB automatically satisfies Eqs. (1), (2), and (3) above, and
therefore eliminates the need to write these equations. As a result,
save yourself some time and effort by always identifying the two-force
members before starting the analysis!
5.5
EXAMPLE
5 .11
The compound beam shown in Fig. 5- 22a is pin connected at B.
Determine the components of reaction at its supports. Neglect its weight
and thickness.
IOkN
(b)
Fig. 5-22
SOLUTION
Free-Body Diagrams. By inspection, if we consider a free-body
diagram of the entire beam ABC, there will be three un!known
reactions at A and one at C. These four unknowns cannot all be
obtained from the three available equations of equilibrium, and so for
the solution it will become necessary to dismember the beam into its
two segments, as shown in Fig. 5-22b.
Equations of Equilibrium. The six unknowns are determined as
follows:
Segment BC
~ "2.J'.r = O; Bx = 0
C+ "i.Ms
+f "i.F,.
= O; -8 kN(l m) + Cy(2 m) = 0
= O;
By - 8kN +Cy = 0
Segment AB
~ "2. J'.r = O; A x - (lOkN)(~) + Bx = 0
C+!MA
= O;
MA - (10 kN) (~)(2 m) - By(4 m)
+f!F,.
= O;
Ay - (10 kN) (~) - By = 0
=
0
Solving each of these equations successively, using previously
calculated results, we obtain
Ax= 6 kN Ay = 12 kN
Bx= 0
By= 4 kN
Cy= 4 kN
MA = 32kN · m
Ans.
Ans.
FRAMES AND M ACHINES
25 3
254
I
CHAPTER
EXAMPLE
5
STRUCTURAL ANALYSIS
s.12
j
The 500-kg elevator car in Fig. 5- 23a is being hoisted at constant speed by
motor A using the pulley system shown. Determine the force developed in
the two cables.
c
500 (9.81) N
(a)
(b)
Fig. 5-23
SOLUTION
We can solve this problem using the free-body
diagrams of the elevator car and pulley C, Fig. 5- 23b. The tensile forces
developed in the two cables are denoted as T1 and T2 .
Free-Body Diagrams.
For pulley C,
Equations of Equilibrium.
(1)
or
For the elevator car,
371 + 272 - 500(9.81) N
=
0
(2)
Substituting Eq. (1) into Eq. (2) yields
3T1 + 2(2T1)
T1
=
500(9.81) N
-
700.71 N
=
=
0
701 N
Ans.
Substituting this result into Eq. (1 ),
T2
=
2(700.71) N
=
1401 N
=
1.40 kN
Ans.
5.5
-
EXAMPLE
5.13
-
The two planks in Fig. 5- 24a are connected together by cable BC and a
smooth spacer DE. Determine the reactions at the smooth supports A
and F, and also find the force developed in the cable and spacer.
100 lb
.I.
1-2ft-i-2
ft-1-2ft -1
200 lb
(a)
(b)
Fig. 5-24
SOLUTION
Free-Body Diagrams. The free-body diagram of each plank is shown
in Fig. 5- 24b. It is important to apply Newton's third Jaw to the interaction
forces as shown.
Equations of Equilibrium.
For plank AD,
~ + IMA = 0; FvE(6 ft) - F8 c(4
ft) - 100 lb (2 ft) = 0
For plank CF,
~ + IMF = 0;
FvE(4 ft) - F8 c(6 ft)
+ 200 lb (2 ft) = 0
Solving simultaneously,
FvE = 140 lb F8 c = 160 lb
Ans.
Using these results, for plank AD,
+f2£,. = O;
NA + 140Jb - 160Jb - lOOJb = 0
NA = 120Jb
Ans.
And for plank CF,
+f 2£,. = 0;
NF + 160 lb - 140 lb - 200 lb = 0
NF = l80 lb
Ans.
FRAMES AND MACHINES
2 55
256
CHAPTER
5
S TRUCTURAL A NA LYS IS
PRELIMINARY PROBLEM
PS-3. In each case, identify any two-force members, and
then draw the free-body diagrams of each member of the
frame.
60N·m
-r
"\=====:::::::::'.):::==~ B
A ,~
0
1.6
I
800N
1.5 m
i - 2m - l -2m--+1
l
200N
r
1.5 m
A
d_
~
I
I
j
~
1~2m~-2m--J
o)
6m
(a)
(d)
400 N/m
T
A~========:::::::=~B
T
t - 2m -
i - - - -3 m ----+1 1 m
~~i
C~::::::'..J _l
0 .. 25 m
1-tm-
t -
2m - l
Yi
B f-- 1.5 m
200N
(e)
v·-,
lm-J-tm- J - -2m--+<
Bo
_l_
0
(b)
A
2m
r
Tm
1
1.5 m
_j
•>
A
2m
- t- --i
2m
0
B
""-. 0.2 m
1.5
o
c
SOON
.,
c
_..!, 16.
(f)
(c)
Prob. PS-3
400N
-
5.5
257
FRAMES AND M ACHINES
FUNDAMENTAL PROBLEMS
All problem solutions must include FBDs.
. ·5-.3. Determine the force P needed to hold the 60-lb
weight in equilibrium.
. ·5-15. If a 100-N force is applied to the handles of the
pliers, determine the clamping force exerted on the smooth
pipe B and the magnitude of the resultant force that one of
the mem bers exerts on pin A.
100 N
p
1- ---250 m m - - - -1
IOON
Prob. FS-15
Prob. FS-13
. '5-1 . Determine the horizontal and vertical components
of reaction at pin C.
5001b
400lb
8
, ~~.~-----'-~---'-~~. c
• 'S-16. Determine the horizontal and vertical components
of reaction at pin C.
400N
~~~· ·+-'"'
i- ~-
•C
lm
4 fl
1•
1- J ft
7:-"
Q
1m
-3 f t - - 3 ft
Pn
~14
- 3 ft-
l
A
Prob. FS-16
258
CHAPTER
5
S TRUCTURAL A NA LYS IS
PROBLEMS
All problem solutions must include FBDs.
*5- 32. Determine the force P required to hold the 100-lb
weight in equilibrium.
5-34. Determine the force P required to hold the 50-kg
block in equilibrium.
~D
B
A
t8l
p
•
0
p
Prob. 5-34
Prob. 5-32
5- 33. In each case, determine the force P required to
maintain equilibrium of the 100-lb block.
5-35. Determine the force P required to hold the 150-kg
crate in equilibrium.
"
0
p
(a)
(b)
Prob. 5-33
(c)
Prob. 5-35
5.5
*5-36. Determine the reactions at the supports A, C, and E
of the compound beam.
12 kN
3kN/m
18
A
-
c
ID
-
E
3m-l-4m-l~l- 6m -~3mj
259
FRAMES AND MACHINES
5-39. The wall crane supports a load of 700 lb. Determine
the horizontal and vertical components of reaction at the
pins A and D. Also, what is the force in the cable at the
winch W?
*5-40. The wall crane supports a load of 700 lb. Determine
the horizontal and vertical components of reaction at the
pins A and D. Also, what is the force in the cable at the
winch W?The jib ABC has a weight of 100 lb and member BD
has a weight of 40 lb. Each member is uniform and has a center
of gravity at its center.
Prob. 5-36
4 ft
5-37. Determine the resultant force at pins A , B, and Con
the three-member frame.
-
4 ft -->.,--1
1 -4 f t - ---1
A
B
E
800N
w
700Jb
Probs. 5-39/40
2m
5-41. Determine the horizontal and vertical components
of force which the pins at A and B exert on the frame.
400 Nr
/n:.:.1
-r
Prob. 5-37
- - - 2m - - -
4~;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;::!1
c
D
1.5 m
Ll---+l~E
5-38. Determine the reactions at the supports at A , E, and B
of the compound beam.
3m
900 N/m
1.5 m
A
3m -
l Jc
3m
-
ID -
B
E
4m - r3m-l-3m-I
Prob. 5-38
Prob. 5-41
.1
260
C H APT ER
5
S TRUCTURAL A NA LYS IS
5-42. Determine the force in members FD and DB of the
frame. Also, find the horizontal and vertical components
of reaction the pin at C exerts on member ABC and
member EDC.
2m
5-45. The pumping unit is used to recover oil. When the
walking beam ABC is horizontal, the force acting in the
wireline at the well head is 250 lb. Determine the torque M
which must be exerted by the motor in order to overcome this
load. The horse-head C weighs 60 lb and has a center of gravity
at Ge. The walking beam ABC has a weight of 130 lb and a
center of gravity at G 8 , and the counterweight has a weight of
200 lb and a center of gravity at Gw. The pitman, AD, is pin
connected at its ends and has negligible weight.
•D-1I
lm
_l
c
•
l -201--1
Prob. 5-42
5-43. Determine the force that the smooth 20-kg cylinder
exerts on members AB and CDB. Also, what are the
horizontal and vertical components of reaction at pin A?
250 lb
Cr-----..J I
D
lm
_l
A
1-1.Smj,_
£ 2m_j
Prob. 5-45
5-46. Determine the force that the jaws I of the metal
cutters exert on the smooth cable C if 100-N forces are
applied to the handles. The jaws are pinned at E and A ,
and D and B. There is also a pin at F.
Prob. 5-43
*5-44. The three power lines exert the forces shown on the
pin-connected members at joints B, C, and D , which in turn are
pin connected to the poles AH and EG. Determine the force
in the guy cable Al and the pin reaction at the support H.
20 ft
A1-1
125 ft
30 nm
I
I-soft
30
ft~-30 ft-~30 ft-1-30 ft
Prob. 5-44
-
15°
50 ft - I
Prob. 5-46
5.5
5-47. The machine shown is used for forming metal plates. It
consists of two toggles ABC and DEF, which are operated by
the hydraulic cylinder H. The toggles push the movable bar G
forward, pressing the plate p into the cavity. If the force which
the plate exerts on the head is P = 12 kN, determine the
force Fin the hydraulic cylinder when 8 = 30°.
FRAMES AND MACHINES
261
5-49. The pipe cutter is clamped around the pipe P. If the
wheel at A exerts a normal force of FA = 80 N on the pipe,
determine the normal forces of wheels B and Con the pipe.
Also calculate the pin reaction on the wheel at C. The three
wheels each have a radius of? mm and the pipe has an outer
radius of 10 mm.
I
\
lOmm
\
\
P = 12kN
lOmm
I
I
/
I
p
Prob. 5-49
Prob. 5-47
*5-48. Determine the horizontal and vertical components
of force which pin C exerts on member ABC. The 600-N
force is applied to the pin.
-
5-50. Determine the force created in the hydraulic
cylinders EF and AD in order to hold the shovel in
equilibrium. The shovel load has a mass of 1.25 Mg and a
center of gravity at G. All joints are pin connected.
2m
D
3m
l,
A
I
1.5 m
I
B
F
-
600N
300N
Prob. 5-48
Prob. 5-50
262
CHAPTER
5
STRUC TURAL ANALYSIS
5-51. The hydraulic crane is used to lift the 1400-lb load
Determine the force in the hydraulic cylinder AB and the
force in links AC and AD when the load is held in the
position shown.
5-53. If d = 0.75 ft and the spring has an unstretched
length of I f1. determine the force F required for equilibrium.
B
F
D
Prob. 5-53
5-54. If a force of F = 50 lb is applied to the pads at A
and C, determine the smallest dimension d required for
equilibrium if the spring bas an unstretched length of 1 ft.
B
F
F
Prob. 5-51
D
Prob. 5-54
*5-52. Determine force P on the cable if the spring is
compressed 25 nun when the mechanism is in the position
shown.1l1e spring has a stiffness of k = 6 kN /m.
5-55. The skid-steer loader has a mass of 1.18 Mg. and in the
position shown the center of mass is at G1• lf there is a 300-kg
stone in the bucket, with center of mass at G 2 • determine the
reactions or each pair of wheels A and B on the ground and
the force in 1he hydraulic cylinder CD and at the pin £.There
is a similar Iinkage on each side of the loader.
-
l.25m
E
ISO mm
k
0.5 111
0.15 m
- - 1.5 m - - - ! - 0.75 m
Prob.5-52
Prob. 5-55
5.5
*5-56.
Determine the force P on the cable if the spring is
compressed 0.5 in. when the mechanism is in the position
shown. The spring has a stiffness of k = 800 lb/ft.
263
FRAMES ANO MACHINES
5-59. The piston C moves vertica lly between the two
smooth walls. Uthe spring has a stiffness of k = 15 lb/in.,
and is unstretched when 8 = Cf, determine the couple
moment that must be applied to AB to hold the mechanism
in equilibrium when 8 = 30°.
B
c
p
24 in.
k
= 15 lb/in.
Prob.5-59
Prob. 5-56
5-57. The spring has an unstretched length of 0.3 m.
Determine the angle 8 for equilibrium if the uniform bars
each have a mass of 20 kg.
5-58. The spring has an unstretched length of 0.3 m.
D etermine the mass m of each uniform bar if each angle
8 = 30" for equilibrium.
*5-60. The platform scale consists of a combination of third
and first class levers so that the load on one lever becomes the
effort that moves the next lever. Through this arrangement, a
small weight can balance a massive object. If x = 450 mm,
determine the required mass of the counterweight S required
to balance the load L having a mass of 90 kg.
5-6L The platform scale consists of a combination of third
and first class levers so that the load on one lever becomes the
effort that moves the next lever. Through this arrangement, a
small weight can balance a massive object. If x = 450 mm, and
the mass of the counte rweight S is 2 kg. determine the mass of
the load L required to maintain the balance.
100 mm 250 mm lf!SO
mm
.H
J:---_2 m
•
E
•
~F~;..
o
G
D
1
150mm
l-350mm
I
•
• B
l1- x-1
A
Probs. 5-57/58
s
Probs. 5-60/61
264
CHAPTER
5
S TRUCTURAL A NA LYS IS
CHAPTER REVIEW
Simple Thuss
A simple truss consists of triangular
elements connected together by pinned
joints. The forces within its members
can be determined by assuming the
members are all two-force members,
connected concurrently at each joint.
The members are either in tension or
compression, or carry no force.
Roof truss
Method of Joints
B
The method of joints states that if a
truss is in equilibrium, then each of its
joints is also in equilibrium. For a plane
truss, the concurrent force system at
each joint must satisfy force equilibrium.
"'l..F. = 0
·'
"'l..F.y = 0
To obtain a numerical solution for the
forces in the members, select a joint
that has a free-body diagram with at
most two unknown forces and one
known force. (This may require first
finding the reactions at the supports.)
Once a member force is determined, use
its value and apply it to an adjacent joint.
Remember that forces that pull on the
joint are rensile forces , and those that
push on the joint are compressive
forces.
To avoid a simultaneous solution of two
equations, set one of the coordinate axes
along the line of action of one of the
unknown forces and sum forces
perpendicular to this axis. This will allow
a direct solution for the other unknown.
The analysis can also be simplified by
first identifying all the zero-force
members.
~SOON
.
F8 A (tension) i<(s•"'Fsc (compression)
265
CHAPTER REVIEW
a
Method of Sections
A4=.====~==::::::j:==~====~I
E
G
a
2 m- - l -2 m--rl--2m- I
1000 N
=0
kF,, = 0
kF,
kMo =O
If possible, sum forces in a direction that
is perpendicular to two of the three
unknown forces. This will yield a direct
solution for the third force.
+ikF,, = 0
- 1000 N +Fee sin 45° = 0
Fee = 1.41 kN (T)
To simplify the analysis, be sure to
recognize all two-force members. They
have equal but opposite collinear forces
at their ends.
<: +kMc =
- -2m- lOOON
0
1000N(4m) - FcF(2m)
FcF = 2kN (C)
2000N
Frames and Machines
The forces acting at the joints of a frame
or machine can be determined by drawing
the free-body diagrams of each of its
members or parts. The principle of
action-reaction should be carefully
observed when indicating these forces on
the free-body diagram of each adjacent
member or pin. For a coplanar force
system, there are three equilibrium
equations available for each member.
_l
F
Three equations of equilibrium are
available to determine the unknowns.
Frames and machines are structures that
contain one or more multiforce members,
that is, members with three or more forces
or couples acting on them. Frames are
designed to support loads, and machines
transmit and alter the effect of forces.
-I
2m
Sectioned members subjected to pulling
are in rension, and those that are
subjected to pushing are in compression.
Sum moments about the point where the
lines of action of two of the three unknown
forces intersect, so that the third unknown
force can be determined directly.
D
0
The method of sections states that if a truss
is in equilibrium, then each part of the
truss is also in equilibrium. Pass a section
through the truss and the member whose
force is to be determined. Then draw the
free-body diagram of the sectioned part
having the least number of forces on it.
i
c
Multiforce
Two-force
member
member
2000N
B
)
I
I
F ABI
£.....+-) F AB
Action- reaction1
c.T
=0
266
C H APT ER
5
S TRUCTURAL A NA LYS IS
REVIEW PROBLEMS
All problem solutions must include FBDs.
RS-1. Determine the force in each member of the truss
and state if the members are in tension or compression.
RS-3. Determine the force in member GI and GC of the
truss and state if the members are in tension or compression.
lO kN
8 kN
4kN
lOOOlb
1.5 m
~~~~J
-===-
1 - - - 2 m - - - + - - - 2m _ __,
lOOO lb
Prob. RS-1
Prob. RS-3
RS-2. Determine the force in each member of the truss
and state if the members are in tension or compression.
G
*RS-4. Determine the force in members GF, FB , and BC
of the Fink l'russ and state if the members are in tension or
compression.
E
1
6001b
10 ft
IA~---+-:'----~~D
IB
c
-I
- ~
l -10 ft - - f - - 10 ft --1---10 ft
lOOO Jb
Prob. RS-2
D
• : i--:----10 ft
Prob. RS-4
267
REVIEW PROBLEMS
R5-5.
Determine the horizontal and vertical components
of force that the pins A and B exert on the two-member frame.
R5-7.
The three pin-connected members shown in the
rop view support a downward force of 60 lb at G. If only
vertical forces are supported at the connections B, C, E and
pad supports A, D, F, determine the reactions at each pad.
lm
I
c
1
.l
lm
B
A
Prob. RS-5
Prob. RS-7
R5-6.
Determine the horizontal and vertical components
of force that the pins A and C exert on the two-member frame.
*RS-8. Determine the resultant forces at pins B and Con
member ABC of the four-member frame.
500N/m
1 - - - - -5 ft - - - - - 1
2f1-1
150 lb/ft
1 - - - - - -3 n1 - - - -+
r--~~~~~-~~~~--i-~~~~
l~~~------t' o.,__
B -----,-.iC •
3 01
1
4 ft
c
- - r-..,.-=~-------~
400N/m
Prob. R5-6
600N/m
I~
~ ~
£ ~Dl
1--2ft - -i-- - - - - 5 ft - - - - - 1
Prob. RS-8
CHAPTER
6
(©Michael N. Paras/AGE Fotostock/Alamy)
The design of these structural members requires finding their
centroid and ca lcu lating their moment of inertia. In this
chapter we wi ll discuss how this is done.
CENTER OF GRAVITY,
CENTROID, AND
MOMENT OF INERTIA
CHAPTER OBJECTIVES
•
To show how to determine the location of the center of gravity and
centroid for a body of arbitrary shape and for a composite body.
•
To present a method for finding the resu ltant of a general
distributed loading.
•
To show how to determine the moment of inertia of an area.
6.1
CENTER OF GRAVITY AND THE
CENTROID OF A BODY
In t his section we will show how to locate the center of gravity for a body
of arbitrary shape, and then we will show that the centroid of the body
can be found using this same method.
Center of Gravity. A body is composed of an infinite number of
particles of differential size, and so if the body is located within a gravitational
field, then each of these particles will have a weight dW. These weights will
form an approximately parallel force system, and the resultant of this system
is the total weight of the body, which passes through a single point called the
center of gravity, G.
269
270
CHAPTER
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
w
x
(a)
Fig. 6-1
To show how to determine the location of the center of gravity, consider
the rod in Fig. 6-la, where the segment having the weight
is located at
the arbitrary position Using the methods outlined in Sec. 3.8, the total
weight of the rod is the sum oif the weights of all of its particles, that is
dW
x.
W= jdw
The location of the center of gravity, measured from the y axis, is
determined by equating the moment of W about they axis, Fig. 6-lb, to
the sum of the moments of the weights of all its particles about this same
axis. Therefore,
xW = jxdW
jxdW
x=
jdw
In a similar manner, if the body represents a plate, Fig. 6- lb, then a
moment balance about the x and y axes would be required to determine
the location (x, Y) of point G. Finally we can generalize this idea to a
three-dimensional body, Fig. 6-lc, and perform a moment balance about
each of the three axes to locate G for any rotated position of the axes. This
results in the following equations.
jxdW
JydW
y=
x=
jdw
jdw
z
=
jzdW
jdw
where
x, y, z are the coordinates of the center of gravity G.
x,y, zare the coordinates of an arbitrary particle in the body.
(6-1)
6.1
CENTER OF
271
GRAVITY A ND THE C ENTROID OF A BODY
w
x
(b)
(c)
Fig. 6-1 (cont.)
Centroid of a Volume. If the body in Fig. 6-2 is made of the same
mate rial, the n its specific weighty (gamma) will be constant throughout
the body. Therefore, a differential element of volume dV will have a
weight dW = y dV. Substituting this into Eqs. 6-1 and canceling out y ,
we obtain formulas tha t locate the centroid C or geometric center of the
body; namely
•C
dV
J-~~~~~~~~~ y
X=
z
y=
(6-2)
=
x
Fig. 6-2
Since these equations represent a balance of the moments of the volume of
the body, then if the volume possesses two planes of symmetry, its centroid
will lie along the line of intersection of these two planes. For example, the
cone in Fig. 6-3 has a centroid that lies on they axis so that x = z = 0.
To find the location y of the centroid, we can use the second of Eqs. 6-2. Here
a single integration is possible if we choose a differential element represented
by a thin disk having a thickness dy and radius r = z. Its volume is
dV = ,,,., 2 dy = ,,,.z 2 dy and its centroid is at = 0, y = y, = 0.
x
y = )'......
r=z
x
y
z
Fig. 6-3
272
CHAPTER
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
Centroid of an Area. If an area lies in the x- y plane and is bounded
by the curve y = f(x ), as shown in Fig. 6-4a, then its centroid will be in this
plane and can be determined from integrals similar to Eqs. 6-2, namely,
(6-3)
y =
x =
These integrals can be evaluated by performing a single integration if we
use a recwngular strip for the differential area element. For example, if a
vertical strip is used, Fig. 6-4b , the area of the element is dA = y dx, and
its centroid is located at = x and y = y /2. If we consider a horizontal
strip, Fig. 6-4c, then dA = x dy, and its centroid is located at = x/2
and y = y.
x
y
x
y
)'
Y = f(x)
y = f(x)
Y = f(x)
d.YJ:
I
(x,y)
•
1-X~
~J-+---• C
ji
-'-+---+---------'--- X
'---'--L-.L.--'------'---
- x-
(a)
x
.______
(b)
I
ji = y
___._-'--I x
(c)
Fig. 6-4
IMPORTANT POINTS
• The centroid represents the geometric center of a body. This
point coincides with the center of gravity only if the material
composing the body is uniform or homogeneous.
• Formulas used to locate the center of gravity or the centroid
represent a balance between the sum of moments of all
the parts of the system and the moment of the "resultant" for
the system.
6.1
CENTER OF GRAVITY AND THE CENTROID OF A BODY
PROCEDURE FOR ANALYSIS
The center of gravity or centroid of an object or shape can be
determined by single integrations using the following procedure.
Differential Element.
• Select an appropriate coordinate system,specify the coordinate
axes, and then choose a differential element for integration.
• For areas the differential element is generally a rectangle of area
dA, having a finite length and differential width.
• For volumes the differential element can be a circular disk of
volume dV, having a finite radius and differential thickness.
• Locate the element so that it touches the arbitrary point (x, y, z)
on the curve that defines the boundary of the shape.
Size and M oment Arms.
• Express the area dA or volume dV of the element in terms of
the coordinates describing the curve.
x, z
• Express the moment arms
y, for the centroid or center of
gravity of the element in terms of the coordinates describing the
curve.
Integrations.
z
• Substitute the formulations for x, y, and dA or dV into the
appropriate equations (Eqs. 6- 1 through 6-3).
• Express the function in the integrand in terms of the same
variable as the differential thickness of the element.
• The limits of the integral are defined from the two extreme
locations of the element's differential thickness, so that when the
elements are "summed" or the integration performed, the entire
region is covered.'
·some formulas for integration are given in Appendix A.
27 3
274
I
CHAPTER
6
CEN TER O F GRAVITY, CE NTROID, AND MO M E NT OF IN ERTIA
6.1
EXAMPLE
Locate the centroid of the area shown in Fig. 6-5a.
y
SOLUTION I
Differential Element.
1 - - - -X - -
(x, y
• (x, y)
A differential element of thickness dx is shown
in Fig. 6-5a. The element intersects the curve at the arbitrary point (x, y),
and so it has a height y .
Area and Moment Arms. The area of the element is dA = y dx, and
its centroid is located at = x, y = y/2.
Integrations. Applying Eqs. 6-3 and integrating with respect to x yields
-,
x
y
3
ixdA
1 - - -lm- - -1
x =
(a)
1m
1m
1 xydx 1 x dx 0.250
0.333
11m
11m
ydx
x dx
11m(y/2)y 11m(x /2)x dx
1m
1m
1 ydx
1 x dx
-
0
0
-
=
0
=
0.100
0.333
Ans.
0
-
0
2
2
dx
lydA
0
-
=
0.3 m
Ans.
2
ldA
y
0.75 m
2
ldA
y =
=
0
0
SOLUTION II
Differential Element.
y =xi
The differential element of thickness dy is
shown in Fig. 6-5b. The element intersects the curve at the arbitrary
point (x, y), and so it has a length (1 - x).
Areaand MomentArms. Theareaoftheelement isdA = (1 -x) dy,
and its centroid is located at
-,
y
-'--1"""-- - - - - - - ' - ---'---'-- X
1 - - -x-- 1 m
1-<t - x~
x
=
x +
Integrations.
(b)
Fig.6-5
1
(1 ;
x =
we obtain
ldA
-
ldA
-
1;
0
-
x, y
y
=
rt
m
2 }0
(l - y) dy
11m(l - x) dy
1 (1 11my(l - x) dy 11m(y - dy
11m
11m(1 - vY) dy
(1 - x) dy
0
lydA
y =
1
m((l + x)/2)(1 - x) dy
-
=
Applying Eqs. 6- 3 and integrating with respect to y,
1
ixdA
x)
=
t m
0.250
0.333
=
0.75 m
Ans.
Vy)dy
y3f2)
=
0.100
0.333
=
0.3 m
Ans.
0
0
NOTE: Plot these results and notice that they seem reasonable. Also, by
comparison, elements of thickness dx offer a simpler solution.
6.1
-
EXAMPLE
6.2
27 5
CENTER OF GRAVITY AND THE CENTROID OF A BODY
-
Locate the centroid of the semi-elliptical area shown in Fig. 6--6a.
y
y
2
- ~=x -1
1 ft
~
dy
f+l=l
(- x,y)
y
x
x
y=y
I
x
-11-dx
- - -2 ft- - - 1 - - -2 ft - - -
2ft
2ft- l
(a)
(b}
Fig. 6-6
SOLUTION I
The rectangular differential element parallel to
they axis shown shaded in Fig. 6--6a will be considered. This element has a
thickness of dx and a height of y.
Area and Moment Arms. Thus, the area is dA = y dx, and its centroid
is located at = x and y = y /2.
Integration. Since the area is symmetrical about they axis,
Differential Element.
x
~ = 0
Applying the second of Eqs. 6-3 with y = ) 1
!f2f (1 -
1
f2f ~(ydx)
{AydA
}A
y =
idA -
- 2ft2
J2fty dx
1
2 - 2ft
- J2ft)1
- 2ft
- 2ft
Am
2
: , we have
2
x )dx
4
4/3
x2 dx = -:;- = 0.424 ft
Ans.
4
SOLUTION II
The shaded rectangular differential element of
thickness dy and length 2x will be considered, Fig. 6--6b.
Area and Moment Arms. The area is dA = 2x dy, and its centroid is at
= Dandy = y.
Integration. Applying the second of Eqs. 6- 3, with x = 2~,
we have
Differential Element.
x
11fty(2x dy)
iydA
y =
-
idA
0
11
ft
0
2xdy
11f• v'1="l
11
0
-
4y
ft
0
+y2=1
dy
4v'l="ldy
4/3
= - ft 'TT
0.424 ft
Ans.
x
276
I
CHAPTER
EXAMPLE
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
6.3
Locate they centroid for the paraboloid of revolution, shown in Fig. 6-7.
- y=
y-
z2
= lOOy
dy
x
1 - -100 mm - - 1
Fig. 6-7
SOLUTION
An element having the shape of a thin disk is
chosen. This element has a thickness dy, it intersects the curve at the
arbitrary point (0, y, z), and so its radius is r = z.
Differential Element.
Volume and Moment Arm. The volume of the element ts
dV = ('1Tz 2) dy, and its centroid is located at y = y.
Applying the second of Eqs. 6-2 and integrating with
respect toy yields
Integration.
fvydV
-
y =
fvdv
1100mm
0 y(1TZ2) dy
1 100mm
o
(1TZ2) dy
1 100mm
1007T 0
y dy
1 100mm =
1007T 0
y dy
2
66.7mm
Ans.
6.1
CENTER OF
GRAVITY AND THE CENTROID OF A BODY
PRELIMINARY PROBLEM
P6-1. lo each case. use the element shown and specify
x.y.and dA.
y
y
I
=x
T
. . . . < - - - - -- - - '
Im
..__ __,_..___ _
__,__~_
1 - --
lm---
(b)
(a)
y
y
y =.r2
7
y = .r
lm
I
Im
' - - - -- ----'--'--X
x
1- - - lm - --
T
l
T
lm
x
l m
l
x
I
Im
(d)
(c)
Prob. P6-1
27 7
278
CHAPTER
6
CEN TER O F GRAVITY, CENTROID, AND MOMENT OF IN ERTIA
FUNDAMENTAL PROBLEMS
F6-1. Determine the centroid (x, Y) of the area.
y
Locate the center of gravity x of the straight rod if its
weight per unit length is given by W = W0 (1 + x2 / L2 ).
F6-4.
)'
........-~ x
~=========-
1 -L- I
lm
y
=x'
-'-----1--"''----------'---x
Prob. F6-4
Prob. F6-1
F6-S. Locate the centroid y of the homogeneous solid
formed by revolving the shaded area about they axis.
1-tm~
F6-2. Determine the centroid (x, Y) of the area.
z
z2 =
)'
0.5
01
.t3
)' =
lm
.!.,,
4-
x
-'-----1-""'-- - - - - - - - - ' - -x
Prob. F6-S
Prob. F6-2
F6-3. Determine the centroid y of the area.
z
Locate the centroid of the homogeneous solid
formed by revolving the shaded area about the z axis.
FIH>.
y
z=
2m
j (12 -
2 ft
--L'- - F - - - --1------4- - -y
-
tm
-lm~
Prob. F6-3
- 1.5
ft~
Prob. FIH>
8y)
6.1
CENTER OF GRAVITY AND THE CENTROID OF A BODY
279
PROBLEMS
6-L Locate the centroid x of the area.
6-5. Locate the centroid x of the area.
6-6. Locate the centroid y of the area.
y
y
h
h
Y=
ax2
1- - -b- - -
1--- -b·- - - 1
Prob. 6-1
6-2. Locate the centroid of the area.
Probs. 6-5/6
6-7. Locate the centroid x of the area.
*6-8.
y
Locate the centroid y of the area.
y
y=4 - 116x2
,v =a cos7TX
L
1
4m
a
~-+-----~~-x
i--sm- 1
-'---'-- - --+- - - - ' - - - x
~ -I
Prob. 6-2
6-3. Locate the centroid x of the area.
*6-4. Locate the centroid y of the area.
Probs. 6-7/8
6-9. Locate the centroid x of the area. Solve the problem
by evaluating the integrals using Simpson's rule.
6-10. Locate the centroid y of the area. Solve the problem
by evaluating the integrals using Simpson's rule.
y
y
y
,
=a.sir'
'
4m
," = .!_
4 x2
'---":::_-------'----x
- - - 4m - - - 1
Probs. 6-3/4
>--------~--- x
1-lm~
Probs. 6-9/10
280
6-11.
CHAPTER
6
CENTER OF GRAV ITY, CE NTR OID, A N D MO M E NT OF INERT IA
6-15.
Locate the centroid ji of the area.
Locate the centroid x of the area.
*6-16. Locate the centroid ji of the area.
y
y
-l
16 ft
4 in.
T
4 rt
1----Sin.~
.__ __,_..__ _ x
l-4rt-j
P rob. 6-U
P robs. 6-15/16
6-17. Locate the centroid
*6-U . Locate the centroid x of the area.
6-18.
6-13. Loca te the centroid ji of th e area.
x of the a rea.
Locate the centroid ji of the area.
y
y
T
"..
Y = h - - r•
a
"
i---a---1
--'--1----------~~-~x
Probs. 6-17/18
6-19. The plate has a thickness of 0.25 ft and a specific
weight of 1' = 180 lb/ ft 3 . Determine the location of its
center of gravity. Also, fmd the tension in each of the cords
used to support it.
P robs. 6-12/13
•
6-14. Locate the centroid ji of the area.
I
y
Y = a/1 x"
"
T
B
I
16 ft
t
yl + /i = 4
1~----0-----1
Pro b. 6-14
P rob. 6-19
--y
c
6.1
CENTER OF GRAVITY AND THE C ENTROID OF A BODY
*6-20. Locate the centroid x of the area.
6-26.
Locate the centroid x of the area.
6-21. Locate the centroid y of the area.
6-27.
Locate the centroid y of the area.
y
281
y
y =a sin f
-I
a
4 ft
Probs. 6-26127
*6-28. The steel plate is 0.3 m thick and has a density of
7850 kg/ m3. Determine the location of its center of gravity.
Also find the reactions at the pin and roller support.
-
- - - 4 ft - - - - - 1
y
6-22.
Probs. 6-20/21
Loca te the centroid x of the area.
6-23.
Locate the centroid y of the area.
y2 = 2x
2m
y
2m
y= - x
B
1-- - 2 m - --'
*6-24.
6-25.
Probs. 6-22123
Locate the centroid x of the area.
Locate the centroid y of the area.
y
Prob.6-28
6-29.
Locate the centroid x of the area.
6-30.
Locate the centroid
y of the area.
y
1- - - - - a - -- - 1
T
"
:y=h - -x"
a"
"
1----~ a -----
Probs. 6-24125
Probs. 6-29/30
282
C HAPT ER
6
CE NTER O F GRAV ITY, CE NTRO I D, AND M OM E NT OF IN ERTIA
z
6-3L Locate the centroid y of the solid.
6-33. Locate the centroid of the solid.
z
l
z
16 in.
x
x
Prob. 6-33
z
1- - -3 ft-----1
6-34. Locate the centroid of the volume.
z
Prob.6-31
*6-32. Locate the centroid of the quarter-cone.
x
z
Prob. 6-34
I
6-35. Locate the centroid of the ellipsoid of revolution.
z
x
x
/
)'
)'
Prob.6-32
Prob. 6-35
6.2
6.2
COMPOSITE BODIES
COMPOSITE BODIES
A composite body consists of a series of connected "simpler" shaped bodies,
which may be rectangular, triangular, semicircular, etc. Such a body, as shown
in Fig. 6--8, can often be divided up into its composite parts and, provided the
weight and location of the center of gravity of each of these parts are known,
we can then eliminate the need for integration to determine the center of
gravity for the body. The method for doing this follows the same procedure
outlined in Sec. 6.1, and so formulas analogous to Eqs. 6-1 result. Here,
however, we have a finite number of weights, and so the equations become
x=
IxW
IW
_
IyW
y = IW
(~)
where
x, y, z
x, y, z
I W
represent the coordinates of the center of gravity
G of the composite body
represent the coordinates of the center of gravity
of each composite part of the body
is the s um of the weights of all the composite
parts of the body, o r simply the total weight of
the body
When the body has a constant density or specific weight, the center of
gravity coincides with the centroid of the body. The centroid for composite
Jines, areas, and volumes can then be found using relations analogous to
Eqs. ~; however, the Ws are replaced by Ls, As, and Vs, respectively.
Centroids for areas and volumes that often make up a composite body
are given in Appendix B.
In order 1ode1ermine lhe force required
to tip over this concrele barrier, it is
[irst necessary to determine the location
of ils cenler of gravity G. This point will
lie on the vertical axis of symmetry.
Fig. 6-8
283
284
CHAPTER
6
CEN TER O F GRAVITY, CENTROID, AND MOMENT OF IN ERTIA
PROCEDURE FOR ANALYSIS
The location of the center of gravity of a body or the centroid of a
composite geometrical object represented by an area or volume can
be determined using the following procedure.
Composite Parts.
• Using a sketch, divide the body or object into a finite number
of composite parts that have simpler shapes.
• If a composite body has a hole, or a geometric region having
no material, then consider the composite body without the
hole and consider the hole as an additional composite part
having negative weight or size.
Moment Arms.
• Establish the coordinate axes on the sketch and determine the
coordinates y, of the center of gravity or centroid of each part.
x, z
Summations.
z
• D etermine x, y, by applying the center of gravity equations,
Eqs. 6-4, or the analogous centroid equations.
• If an object is symmetrical about an axis, the centroid of the
object lies on this axis.
If desired, the calculations can be arranged in tabular form, as
indicated in the following examples.
):. a
. ~tv~.
..•
The center of gravity of this water tank can be
determined by dividing it into composite parts and
applying Eqs. 6-4.
•
.,
6.2
EXAMPLE
COMPOSITE BODIES
285
6 .4
Locate the centroid of the plate area shown in Fig. 6-9a.
y
2 ft
-f
lft
~~----,,.----1-----~-x
hr1l- 2 r t -- -3 ft- I
(a)
Fig. 6-9
SOLUTION
y
Composite Parts. The plate is divided into three segments as shown
in Fig. 6-9b. Here the area of the small rectangle is considered "negative"
since it must be subtracted from the larger area * ·
, .I
Moment Arms. The location of the centroid of each segment is shown
in the figure. Note that the coordinates of* and are negative.
x
1.5 ft
Summations. Taking the data from Fig. 6-9b, the calculations are -~~i::~::i:::::;::r--~-x
15 ft I rt
tabulated as follows:
I
Segment
A (ft 2)
x (ft)
y (ft)
,{A (ft3)
l
= 4.5
(3)(3) = 9
-(2)(1) = -2
IA = 11.5
1
1
4.5
4.5
- 13.5
13.5
5
-4
IyA = 14
2
3
4'3)(3)
-1.5
-2.5
1.5
2
IxA
=-4
y
0
Thus,
m2.sft-
ITT
2 ft
---~-~----- x
:X
_
y
=
=
IxA
IA
IyA
!A
=
=
-4
ll.5
14
11.5
(b)
= -0.348 ft
Ans.
= 1.22 ft
Ans.
NOTE: If these results are plotted in Fig. 6-9a, the location of point C
seems reasonable.
286
I
CHAPTER
EXAMPLE
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
6.5
z
Locate the center of gravity of the assembly shown in Fig. 6-lOa. The
conical frustum has a density of Pc = 8 Mg/m3, and the hemisphere has
a density of Plr = 4 Mg/m3. There is a 25-mm-radius cylindrical hole in
the center of the frustum.
SOLUTION
lOOmm
50 mm.--a~~..--1----y
50mm
I
x
(a)
Fig. 6-10
The assembly can be thought of as consisting of
four segments as shown in Fig. 6-lOb. For the calculations, @ and©
must be considered as " negative" segments in order that the four
segments, when added together, yield the total composite shape
shown in Fig. 6-lOa.
Composite Parts.
Using the table in Appendix B, the calculations for
the centroid of each piece are shown in the figure.
Summations. Because of symmetry,
Moment Arm.
z
x= y=O
Ans.
Since W = mg, and g is constant, the third of Eqs. 6-4 becomes
= Izmf:£m. The mass of each piece can be calculated from m = pV.
Also, 1 Mg/m3 = 10- 6 kg/mm3, so that
z
Segment
z (mm)
m (kg)
zm (kg·mm)
1
8(10-6) ( ~) 7T(50) 2(200)
=
4.189
50
209.440
2
4(10- 6) ( ~) 7T(50)3
=
1.047
-18.75
-19.635
3
-8(10- 6) ( ~) 7T(25)2(100)
=
-0.524
100 + 25
4
-8(10-6)7T(25)2 (100)
=
-1.571
50
=
-65.450
125
-78.540
Izm
Im = 3.142
Thus,
z=
Izm
Im
=
45.815
3.142
=
14. 6 mm
l_
200mm
100 mm - · 25 mm
4
-,
0
(b)
=
45.815
Ans.
I
25mm
J
ti
lOOmm
6.2
287
COMPOSITE BODIES
FUNDAMENTAL PROBLEMS
• ~7. Locate the cenlroid (x. y, Z) of the wire bent in the
shape shown.
• 6-1 .
Locate the cen troid (x. Y) of the cross-sectional
area.
y
0.5 in.
x
600mm
y
~
I
400 mm
y
Prob.
1'6-8.
Locate the centroid
0.5 in.
x
~6-7
y of the
Proh. F6-IO
beam's cross-sectional
Locate the center of gravity (x, y, z) of the
homogeneous solid block.
area.
F6-ll.
y
z
150 mm ISO mm
I
y
I
mm
Pr•
• r.-~
Locate the centroid
F6-8
~ r•
y of Lhe beam's cross-sectional
area.
ti. l "-•-
116-U. Locate the center of gravity (x. y, Z) of the
homogeneous solid block.
1- - - 400 mm - - -
c
200mm
D-
1.8 Ill
---;.__/ -Ly
.sm/
Pr."'. ...6-9
Proh.16- 2
288
CHAPTER
6
CEN TER OF GRAVITY, CEN TRO ID, AND MO M EN T OF INERTIA
PROBLEMS
*6-36. Locate the centroid (x, y) of the area.
,,
6-39. Locate the centroid (x, ji) of the area.
y
·1- 6in.- I
l - 6in.-J
l in.
3 in .
6
Tf------+----ll_X
..L
I
l
6 in.
6 in.
-~~----•-------- x
_I .__________._______,
l -6in.- I
l - 6in.- I
Prob. 6-36
6-37. Locate the centroid
area.
y for the beam's cross-sectional
Prob. 6-39
*6-40. Locate the centroid y of the beam's cross-sectional
area. Neglect the size of the corner welds at A and B for the
calculation.
r
llOmm
y
Jc
-
240mm
120 mm
I
I
15 mm
240mm
y
Prob. 6-37
6-38. Locate the centroid
sectional area shown.
y of the beam having the crossProb. 6-40
_i_ 1 - -150 mm - -11
15
6-41. Locate the centroid (x, ji) of the area.
mm 1~------~1 T
r ,_:_c~l
y
x
150mm
15mm I
I
- 15mm
A
3in.
I
l - 1oomm - - I
Prob. 6-38
1 in.
1-3in.-1-3in.
Prob. 6-41
6.2
6-42.
Locate the centroid (x, y) of the area.
COMPOSITE BODIES
289
6-45. A triangular plate made of homogeneous material
has a constant thickness that is very small. If it is folded over
as shown, determine the location y of the plate's center of
gravity G.
y
l,,;,l
-,
1.5 in.
T
l_
~
I.Sin.
x
~1.5 in.~
Prob. 6-42
6 in.
6-43. Locate the centroid y of the cross-sectional area of
the beam. The beam is symmetric with respect to the y axis.
y
x
Prob. 6-45
Ty
1
3 in.
c
J
1 in.
x
6-46. A triangular plate made of homogeneous material
has a constant thickness that is very small. If it is folded over
as shown, determine the location of the plate's center of
gravity G.
z
Prob. 6-43
*6-44. Locate the centroid y of the cross-sectional area of
the beam constructed from a channel and a plate. Assume
all comers are square and neglect the size of the weld at A.
20 m
y
350mm
c
6 in.
lOmm
x
- - 325 mm - --i---325 mm - -
Prob. 6-44
~
............... 3.m.
Prob. 6-46
290
CHAPTER
6
CEN TER OF GRAVITY, CEN TRO ID, AND MO M EN T OF INERTIA
6-47. The assembly is made from a steel hemisphere,
P.i = 7.80 Mg/m3 ,
and
an
aluminum
cylinder,
Pai = 2.70 Mg/m3. Determine the center of gravity of the
assembly if the height of the cylinder is h = 200 mm.
6-50. Determine thedistanceh to which a 100-mrn-diameter
hole must be bored into the base of the cone so that the
center of gravity of the resulting shape is located at
= 115 mm. The material has a density of 8 Mg/m3.
z
*6-48. The assembly is made from a steel hemisphere,
= 7.80 Mg/m3 ,
and
an
aluminum
cylinder,
Pal = 2.70 Mg/m . Determine the height h of the cylinder
so that the center of gravity of the assembly is located at
160mm.
P.i
3
z
z=
z
80mm
500mm
h
l
r
160mm
x
x
Probs. 6-47/48
6-49. The car rests on four scales and in this position the
scale readings of both the front and rear tires are shown by
FA and F8 . When the rear wheels are elevated to a height of
3 ft above the front scales, the new readings of the front
wheels are also recorded. Use this data to calculate the
location x and y to the center of gravity G of the car. The
tires each have a diameter of 1.98 ft.
Prob. 6-50
z
6-51. Determine the distance to the centroid of the shape
that consists of a cone with a hole of height h = 50 mm
bored into its base.
z
1
FA = 1129 lb + 1168 lb = 2297 lb
F8 = 975 lb+ 984 lb = 1959 lb
(
FA = 1269 lb + 1307 lb = 2576 lb
Prob. 6-49
500mm
x
Prob. 6-51
6.2
z
*6-52. Locate the center of gravity of the assembly. The
cylinder and the cone are made from materials having
densities of S Mgtm3 and 9 Mglm3, respectively.
291
COMPOSITE BODIES
6-54. The assembly consists of a 20-in. wooden dowel rod
and a tight-fitting steel collar. Determine the distance 'i to
its center of gravity if the specific weights of the materials
are 1'w = 150 lb/ft 3 and y., = 490 lb/ft 3. The radii of the
dowel and collar are shown.
z
I
·-
- ,.-
--- -
-5 in.
10 in._
I
G
\..
X-
0.4 m
l
/•
/
x
7
2in.
1 m.
Prob. 6-54
0.2 111
Prob. 6-52
6-55. The composite plate is made from both steel (A)
and brass (B) segments. Determine the weight and location
('i, y, Z) of its center of gravity G. Take Pst = 7.85 Mg/m3,
and Pbr = 8.74 Mg/m3.
6-53. Major floor loadings in a shop are caused by the
weights of the objects shown. Each force acts through its
respective center of gravity G. Locate the center of gravity
(x, Y) of all these components.
I
y
450lb
225mm
y
Prob. 6-53
Prob. 6-55
292
CHAPTER
6
CEN TER O F GRAVITY, CE NTROID, AND MO M E NT OF IN ERTIA
6.3
MOMENTS OF INERTIA FOR
AREAS
In the first few sections of this chapter, we determined the centroid for an
area by considering the first moment of the area about an axis; that is, for
the computation we had to evaluate an integral of the form x dA. An
2
integral of the second moment of an area, such as x dA , is referred to as
the moment of inertia for the area. Integrals of this form arise in formulas
used in mechanics of materials, and so we should become familiar with the
methods used for their computation.
J
y
Moment of Inertia. Con.sider the area A, shown in Fig. 6- 11, which
--x
lies in the x - y plane. By definition, the moments of inertia of the
differential planar area dA about the x and y axes are df, = y2 dA and
dly = x 2dA, respectively. For the entire area the moments of inertia are
determined by integration; i.e.,
dA
r
0
J
/T
y
x
Fig. 6-11
(6-5)
We can also formulate this quantity for dA about the "pole" 0 or
z axis, Fig. 6-11. This is referred to as the polar moment of inertia. It is
defined as dl0 = r 2dA , wheire r is the perpendicular distance from the
pole (z axis) to the element dA . For the entire area the polar moment of
inertia is
(6-6)
Notice that this relation between 10 and I., ly is possible smce
r 2 = x 2 + y 2 , Fig. 6- 11.
From the above formulations it is seen that I., />" and 10 will always be
positive since they involve t!he product of distance squared and area.
Furthermore, the units for moment of inertia involve length raised to the
4
. 4
f ourth power, e.g., m4, mm,
oir ft4, m
.
6.4
6.4
293
PARAllEL-AxlS THEOREM FOR AN AREA
PARALLEL-AXIS THEOREM
FOR AN AREA
If the moment of inertia is known about an axis passing through the
centroid of an area, then the parallel-axis theorem can be used to find
the moment of inertia of the area about any axis that is parallel to the
y
y'
x'
centroidal axis. To develop this theorem, consider finding the moment
of inertia about the x axis of the shaded area shown in Fig. 6-12.
If we choose a differe ntial element dA located at an arbitrary
distance y' from the centroida/ x ' axis, then the distance between the
parallel x and x ' axes is d,,, and so the moment of inertia of dA about
the x axis is dlx = (y' + d,,)2 dA. For the entire area,
Ix
=
i
I
y'
I
x·
2
(y' + d,,) dA
2
= iy' dA
+ 2d,, ly' dA +
" - - - - - - - ' - -- - -- x
d~ idA
0
Fig. 6-12
The first integra l represents the moment of inertia of the area about
the centroidal axis, 7.... The second integral is zero since the x' axis
passes through the area's centroid C; i.e., y' dA = Y' dA = 0 since
y• = 0. Since the third integral represents the area A, the final result
is therefore
J
J
(6-7)
A similar expression can be written for I,,; i.e.,
(6-8)
And finally, for the polar moment of inertia, since l e
2 we have
d 2 = d x2 + dY'
I lo =le + Ad21
=
Ix· + 1,,. and
(6-9)
The form of each of these three equations states that the monwnt of
inertia for an area abow an axis is equal to its moment of inertia about a
parallel axis passing through the area's centroid, plus the product of the
area and the square of th e perpendicular distance between the axes.
294
CHAPTER
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
PROCEDURE FOR ANALYSIS
y
y = f(x)
dA
dyI
l---------r:----1·
(x,y)
1---x - -
y
~-------~~- x
(a)
In most cases the moment of inertia can be determined using a single
integration. The following procedure shows two ways in which this
can be done.
• If the curve defining the boundary of the area is expressed as
y = j(x), then select a rectangular differential element such
that it has a finite length and differential width.
• The element should be [ocated so that it intersects the curve at
the arbitrary point (x, y ).
y
Case 1:
- x-j
""'l/(x,y)
T
Y = f(x)
y
dA
• Orient the element so that its length is parallel to the axis
about which the moment of inertia is calculated. This situation
occurs when the rectangular element shown in Fig. 6-13a is
used to determine I.r for the area. Here the entire element is
at a distance y from the x axis since it has a thickness dy. Thus
I, =
y 2 dA. To find ly, the element is oriented as shown in
Fig. 6-13b. This element lies at the same distance x from the
y axis so that ly =
x 2 dA.
J
~-~~----~-- x
~
(b)
J
Case 2:
Fig. 6-13
• The length of the element can be oriented perpendicular to the
axis about which the moment of inertia is calculated; however,
Eq. 6-7 does not apply since all points on the element will not
lie at the same moment-arm distance from the axis. For example,
if the rectangular element in Fig. 6-13a is used to determine f,.,
it will first be necessary to calculate the moment of inertia of
the element about an axis parallel to the y axis that passes
through the element 's centroid, and then determine
the moment of inertia of the element about they axis using the
parallel-axis theorem. Integration of this result will yield ly. The
details are given in Example 6-7.
6.4
I
EXAMPLE
6 .6
y'
Detennine the moment of inertia for the rectangular area shown in
Fig. Crl4 with respect to (a) the centroidal x' axis, (b) the axis XfJ passing
through the base of the rectangle, and ( c) the pole or z' axis perpendicular
2
to the x ' -y' plane and passing through the centroid C.
T
,,
SOLUTION (CASE 1)
Part (a ). The differential element shown in Fig. Cr14 is chosen for
integration. Because of its location and orientation, the entire element
is at a distance y' from the x ' axis. Here it is necessary to integrate
from y' = -h/2 toy' = h/2. Since dA = b dy', then
fx· =
295
PARALLEL-AxlS THEOREM FOR AN AREA
J
=
y'2 dA
A
f
lr/2
y'2 (b dy')
- lr/2
=b
J"/2
y'2 dy'
1
3
/x. = -12
bh
Ans.
Part (b). The moment of inertia about an axis passing through the
base of the rectangle can be obtained by using the above result and
applying the parallel-axis theorem, Eq. Cr7.
-
-
l x• - Ix·
+ Ady2
(h)
1 3 + bh = -bh
12
2
2
= -1 bh3
3
Ans.
Part (c). To obtain the polar moment of inertia about point C, we
must first obtain ly'• which may be found by interchanging the
dimensions band h in the result of part (a), i.e.,
1·
y
I
y'
c
,,
2
l
l-~-1-~ -I
Fi g. 6-14
- lr/2
-
t
dy'
_L
1 3
= -hb
12
Using Eq. Cr9, the polar moment of inertia about C is therefore
1
2
2
l e = I x' + ly' =
bh(h + b )
Ans.
12
I
:c'
296
CHAPTER
EXAMPLE
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
6. 7
-
-
Determine the moment of inertia of the shaded area shown m
Fig. 6-15a about the x axis.
y
v2 = 400x
-~
SOLUTION I {CASE 1)
-,
(100 - x)
dy
- - - - - ----< _l._
-,
-
200mm
y
-'~~x
1-lOOmm-I
(a)
y
y2 = 400x
~
T
l =-x~l:~x~
Y
200mm
- - - + - - - - + - I - X'
_I y
y= z-
A differential element that is parallel to the x axis, Fig. 6-15a, is chosen
for integration. It intersects tbe curve at the arbitrary point (x, y).
Since this element has a thickness dy and intersects the curve at the
arbitrary point (x,y), its area is dA = (100 - x) dy. Furthermore, the
element lies at the same distance y from the x axis. Hence, integrating
with respect toy, from y = 0 toy = 200 mm, we have
fo oommy2(100 - x)dy
2
I, = irdA =
=
12oommy\100 -
=
107(106) mm4
12oomm(100y2 -
:~) dy
Ans.
SOLUTION II {CASE 2)
A differential element parallel to they axis, Fig. 6-15b, is chosen for
integration. It intersects the curve at the arbitrary point (x, y). In
this case, all points of the element do not lie at the same distance
from the x axis, and therefore the parallel-axis theorem must be
used to determine the moment of inertia of the element with respect
to this axis. For a rectangle having a baseband height h, the moment
of inertia about its centroidal axis has been determined in part (a)
of Example 6-6. There it was found that lt' = 112 bh 3. For the
differential element shown in Fig. 6-15b, b = dx and h = y, and so
dl_,. = 112 dxy3 • Since the centroid of the element is y = y/2 from
the x axis, the moment of inertia of the element about this axis is
'----------'--- x
(b)
:~) dy =
df., = dl_t'
+ dA
r 1~ dx y
=
3
+ y dx (~)
2
=
~y3 dx
(This result can also be concluded from part (b) of Example 6-6.)
Integrating with respect to x, from x = 0 to x = 100 mm, yields
Fig. 6-15
/., =
=
J
df., =
rlOOmml
3
3y dx
Jo
107(106) mm4
rtOOmml
=
}
0
3
3 2
( 400x) 1 dx
Ans.
6.4
PARALLEL-Axis THEOREM FOR AN AREA
297
FUNDAMENTAL PROBLEMS
F6-13. Determine the moment of inertia of the area about
the x axis.
F6-15. Determine the moment of inertia of the area about
they axis.
y
)'
y3 = x2
lm
r
y3 = x2
lm
1---lm ~
1 - - -lm - - -
Prob.F6-13
Prob.F6-15
F6-14. Determine the moment of inertia of the area about
the x axis.
F6-16. Determine the moment of inertia of the area about
they axis.
y
)'
r
lm
lm
~-+---------~x
1 - - -lm - - -
Prob.F6-14
~-+-----------.X
1---lm ~
Prob.F6-16
298
CHAPTER
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
PROBLEMS
*6-56. Determine the moment of inertia of the area about
the x axis.
*6-60. Determine the moment of inertia for the area
about the x axis.
6-57. Determine the moment of inertia of the area about
they axis.
6-61. Determine the moment of inertia for the area about
they axis.
y
y
y =x'f2
Y=
1
.!l... x"
ti'
b
lm
1- - - - a - - - -
---+----------+~---x
Probs. 6-56/57
i - - - - -1 m - - -...
Probs. 6-60/61
6-58. Determine the moment of inertia for the area about
the x axis.
6-59. Determine the moment of inertia for the area about
they axis.
6-62. Determine the moment of inertia for the area about
the x axis.
6-63. Determine the moment of inertia for the area about
they axis.
y
-
lOOmm -1
)I
200mm
,"=...1...
50 x2
-1
y2 = 1 -
0.5x
lm
--+--+-----------~-- x
1 - - - - - 201 - - - - -
Probs. 6-58/59
Probs. 6-62/63
6.4
*6-64. Determine the moment of inertia of the area about
the x axis. Solve the problem in two ways, using rectangular
differential elements: (a) having a thickness dx and
(b) having a thickness of dy.
PARALLEL-Axis THEOREM FOR AN AREA
299
*6-68. Determine the moment of inertia about the x axis.
6-69.
Determine the moment of inertia about they axis.
y
y
y = 2.5 - O.lx2
I
2.5 ft
--'--'------+------.,,_-X
t -- 5tt- I
x2 + 4y2 =
T
4
lm
Prob.6-64
~--+-----------~--x
,___ _ _ _ 2 m _ _ _ __,
6-65. Determine the moment of inertia of the area about
the x axis.
6-70. Determine the moment of inertia for the area about
the x axis.
y
•.2
y
Probs. 6-68169
= lt2 x
b"
h
)'
- --+- - - - - - --1-'---X
1----b---~
Prob. 6-65
-r
4
6-66. Determine the moment of inertia for the area about
thex axis.
in.
l_.__~~~~~-'---X
- - - 16 in. - - -1
1
I
--
Prob.6-70
6-67. Determine the moment of inertia for the area about
they axis.
6-71. Determine the moment of inertia for the area about
they axis.
)'
r
)'
8m
-r
r - - y = _.!_x3
I
s
4
in.
-1.---1-""'-- - ---'-- -x
J_:1=--=--=----_- 1-6-in-.-=--=--=----_--_'-,- -x
Probs. 6-66/67
Prob.6-71
4m-I
300
CHAPTER
6
CEN TER OF GRAVITY, CEN TRO ID, AND MO M EN T OF INERTIA
*6-72. Determine the moment of inertia for the area
about the x axis.
*6-76. Determine the moment of inertia for the area
about the x axis.
6-77. Determine the moment of inertia for the area about
they axis.
y
y = .!!._ x3
b3
T,,
)'
-1
2m
y=x
l- - - - -b - - - - -
Prob.6-72
--+--------!--~- x
1--
2m ----J
Probs. 6-76n7
6-73. Determine the moment of inertia for the area about
they axis.
6-78. Determine the moment of inertia for the area about
the x axis.
y
T,,
y
b
v=bx2
,
l- - - - -b - - - - -
a2
Prob.6-73
Prob.6-78
6-74. Determine the moment of inertia for the area about
the x axis.
6-75. Determine the moment of inertia for the area about
they axis.
6-79. Determine the moment of inertia for the area about
they axis.
y
y
b2
a
y2 = - x
. ,rHr----1,-1--
lm
x
b
,, =bx2
,
a2
_l__ l m
1---- a ---~
Probs. 6-74ns
Prob.6-79
6.5
6. 5
MOMENTS OF INERTIA FOR C OM POSITE A REAS
MOMENTS OF INERTIA FOR
COMPOSITE AREAS
The moment of inertia of a composite area that consists of a series of
connected "simpler" parts or shapes can be determined about any axis
provided the moment of inertia of each of its parts is known or can be
determined about the axis. The following procedure outlines a method for
doing this.
PROCEDURE FOR ANAL YS/S
Composit e Parts.
• Using a sketch, divide the area into its composite parts and
indicate the perpendicular distance from the centroid of each
part to the axis.
Paralle l-Axis Theorem.
• If the centroidal axis for each part does not coincide with the
axis, the parallel-axis theorem, I = l + Ad2 , must be used to
determine the moment of inertia of the part about the axis. For
the calculation of 7, use Appendix B.
Summat ion.
• The moment of inertia of the entire area about the axis is
determined by summing the results of its composite parts
about this axis.
• If a composite part has an empty region or hole, its moment of
inertia is found by subtracting the moment of inertia of the hole
from the moment of inertia of the entire part including the hole.
To design or analyze 1his T-beam, ii is
necessary 10 locale the centroidal axis
of its cross-sectional area. a nd Lhen find
lhe mo ment of inerti a of the area about
lhis axis.
30 1
302
I
CHAPTER
EX AMPLE
6
CENTER OF GRAVITY, CENTROID, AND MOMENT OF INERTIA
6.8
Determine the moment of inertia of the area shown in Fig. 6-16a
about the x axis.
1 - 1000101 - I
1 -1000101 - I
T
T
750101
250101
250101
750101
-8
--+-+---•
750101
750101
I
-'--~----~--
(a)
x
(b)
Fig. 6-16
SOLUTION
The area can be obtained by subtracting the
circle from the rectangle shown in Fig. 6-16b. The centroid of each
area is shown in the figure.
Parallel-Axis Theorem. The moments of inertia about the x axis
are determined using the parallel-axis theorem and the moment of
inertia formulas for circular and rectangular areas, I, = l7Tr 4 and
I, = 1\bh3 , found in Appendix B.
Composite Parts.
Circle
z, = Zr.' + Ad,.
-
=
2
1
-4 7T(25) 4 + 7T(25) 2(75)2 = 11 .4(10 6) mm4
Rectangle
z, = Zr.' + Ady
-
=
?
1
(100)(150)3
12
+
(100)(150)(75) 2 = 112.5(106) mm4
Summation. The moment of inertia for the area is therefore
I,
=
-11.4(106)
+ 112.5(106)
=
101(106) mm4
Ans.
6.5
EXAMPLE
30 3
MOMENTS OF INERTIA FOR COMPOSITE AREAS
6 .9
Determine the moments of inertia for the cross-sectional area of the
member shown in Fig. 6-17a about the x and y centroidal axes.
y
I00 m!!!_j
1-:
I
g
400mm
SOLUTION
Composite Parts. The cross section can be subdivided into the three
rectangular areas A, B, and D shown in Fig. 6-17b. For the calcwation,
the centroid of each of these rectangles is located in the figure.
400-
IOOmm
rr
-I
l-1oomm
Parallel-Axis Theorem. F rom the table in Appendix B, or from
Example 6-6, the moment of inertia of a rectangle about its cen troidal
axis is 1 = ftbh 3. Hence, using the parallel-axis t heorem for rectangles
A and D, the calculations are as follows:
600mm~
(a)
100
Rectangles A and D
Ix =
Zr• +
_
1
200mm
Ad; =
l~ (100)(300)
3
TA
300mm •
2
+ (100)(300)(200)
250mm
...L H""'"-+--+---M
8
- -r---++--- X
I
-r
= 1.425(109) mm4
ly
y
m!!!_I
= l y· + Ad; = l~ (300)(100)3 + (100)(300)(250)2
250mm ' ~~ .
200mm D
(b)
Rectangle B
Fig. 6-17
Ix
= l~ (600)(100)3 = 0.05(109) mm4
ly
= 1~ (100)(600)3 = 1.80(109)
mm4
The moments of inertia for the entire cross section are
ix = 2[1.425(109)] + 0.05(109)
= 2.90(109)
mm4
ly = 2[1.90(109)]
= 5.60(109)
Ans.
+ 1.80(109 )
mm4
_L
-j 1-toomm
= 1.90(109) mm4
Summation.
t hus
300mm
Ans.
304
CHAPTER
6
CEN TER OF GRAVITY, CEN TRO ID, AND MO M EN T OF INERTIA
FUNDAMENTAL PROBLEMS
F6-17. Determine the moment of inertia of the cross·
sectional area of the beam about the centroidal x and y axes.
F6-19. Determine the moment of inertia of the crosssectional area of the channel with respect to they axis.
y
)'
200mm
f
I
50 mm
50mm
:::::::!-+-1---~..,..---x
200mm
I
1150 mnJ
(1so
mm
SO mm
Prob.F6-17
Prob.F6-19
F6-18. Determine the moment of inertia of the crosssectional area of the beam about the centroidal x and y axes.
F6-20. Determine the moment of inertia of the crosssectional area of the T-beam with respect to the x' axis
passing through the centroid of the cross section.
30mm
1-~
)'
0
T
,,,__--+-----1i+ •-"-'+1 30 mm
I
150mm
200mm
0
_I_
__ 30mm
30
1-1-300 mm -1-1
30mm
30mm
Prob.F6-18
---+--+--+-------,.--x·
nu~!--------'f'---YLI
- - l -150 mm - I
Prob.F6-20
6.5
MOMENTS OF INERTIA FOR COMPOSITE AREAS
305
PROBLEMS
*6-80. Determine the moment of inertia of the composite
area about the x axis.
6-81. Determine the moment of inertia of the composite
area about they axis.
6-83. Determine the location y of the centroid of the
cross-sectional area of the channel, and then calculate the
moment of inertia of this area about this axis.
SO mm
SO mm
y
- 3 in.
-1--
6 in.
T3 in.
x
+
3 in.
~l-1----------~--x
Probs. 6-80/81
Prob.6-83
6-82. The polar moment of inertia for the area is
Jc = 642 (106) mm4, about the z' axis passing through the
centroid C. The moment of inertia about the y' axis is
264 (106) mm 4 , and the moment of inertia about the x axis is
938 (106) mm4• Determine the area A.
*6-84. Determine y, which locates the centroidal axis x'
for the cross-sectional area of the T-beam, and then find the
moments of inertia Ix· and I, ..
.Y'
y'
75 ml!!...
150mm
200mm
Prob.6-82
Prob.6-84
306
CHAPTER
6
CEN TER OF GRAVITY, CEN TRO ID, AND MO M EN T OF INERTIA
6-85. Determine the moment of inertia of the cross-sectional
area of the beam about the x axis.
6-89. Determine the moment of inertia of the cross-sectional
area of the beam about they axis.
6-86. Determine the moment of inertia of the cross-sectional
area of the beam about they axis.
6-90. Determine y, which locates the centroidal axis x' for
the cross-sectional area of the T-beam, and then find the
moment of inertia about the x' axis.
y
y
50mm
-
l in.
8 in.
x'
250mm
,__ _ _ _ 10 in. _ _ __,
1 in.
Probs. 6-85/86
Probs. 6-89/90
6-87. Determine the moment of inertia I., of the area about
the x axis.
*6-88. Determine the moment of inertia I, of the area
about the y axis.
6-91. Determine the moment of inertia of the cross-sectional
area of the beam about the x axis.
*6-92. Determine the moment of inertia of the cross-sectional
area of the beam about they axis.
y
y
~ 150mm
150mm ~1
-1
200mm
150mm
t
150mm
- - -+ - - - - - - + - X
75mm
l~--~-x
0
Probs. 6-87/88
Probs. 6-91/92
CHAPTER REVIEW
CHAPTER REVIEW
Center of Gravity and Centroid
The cenrer of gravity G represents a point
where the weight of the body can be
considered concentrated. The distance from
an axis to this point can be determined from
a balance of moments, which requires that
the moment of the weight of all the particles
of the body about this axis must equal the
moment of the entire weight of the body
about the axis.
z
x=
y=
z=
x
The cenrroid is the location of the geometric
center for the body. It is determined in a
similar manner, using a moment balance of
geometric elements such as area or volume
segments. For bodies having an arbitrary
shape, moments are summed (integrated)
using differential elements.
Composite Body
If the body is a composite of several shapes,
each having a known location for its center of
gravity or centroid, then the location of the
center of gravity or centroid of the body can
be determined from a discrete summation
using its composite parts.
x
307
308
C HAPT ER
6
CE NTER O F GRAV ITY, C ENTRO I D, AND MOM E NT OF IN ERTIA
Area Moment of Inertia
y
The area and polar momenis of inerria
represent the second moment of the area
about an axis. It is frequently used in
formulas throughout mechanics of materials.
If the area shape is irregular but can be
described mathematically, then a differential
element must be selected and integration over
the entire area must be performed to determine
the moment of inertia.
A
I,= L/dA
--x -
ly
10
Lx
= Lr
=
2
dA
r
2
dA
1---1
/
T
y
IL__ _ _ _ _ _L __ _ _
x
0
Parallel-Axis Theorem
If the moment of inertia for an area is known
about its centroidal axis, then its moment
of inertia about a parallel axis can be
determined using the parallel-axis theorem.
dA
A
I= I+ Ad2
--1'-----....----+--~~ r
c
d
-------'----!
Composite Area
If an area is a composite of common shapes,
then its moment of inertia is equal to the
algebraic sum of the moments of inertia of
each of its parts.
--1---•
---1----~-- x
REVIEW PROBLEMS
R6-L Locate the centroid x of the area.
R6-2. Locate the centroid y of the area.
R6-3.
Locate the centroid of the rod.
z
y
1
4 ft
)' =
c2
J
--1-~---~- x
a-I
I
1---b- I
Probs. R6-l/2
y
Prob. R6-3
I
REVIEW PROBLEMS
*R6-4. Locate the centroid y of the cross-sectional area of
the beam.
R6-6. Determine the area moment of inertia of the area
about they axis.
y
SO mm
- 75mm
y
75mm-
1=.--ri---t'"
2smm r
100mm
l
C
y
-J 1-
309
L-----'---L.=-----'---x
-J l-
25 mm
25mm
4y = 4 - .~
I
1 ft
__,_1--'------+------'---x
1---- 2 ft
---1
Prob.R6-6
Prob.R6-4
R6-5. Determine the moment of inertia for the area
about the x axis.
R6-7. Determine the area moment of inertia of the
cross-sectional area of the beam about the x axis which
passes through the centroid C.
y
y
2 in.
1 - - - - - 4 in. - - - - -1
Prob.R6-5
1-4
!!
2
_J
I
Prob.R6-7
CHAPTER
7
The bolts used for the connections of this steel framework are subjected to
stress. In this chapter, we will discuss how engineers design these connections
and their fasteners.
STRESS AND
STRAIN
CHAPTER OBJECTIVES
•
To show how to use the method of sections for determining the
internal loadings in a member.
•
To introduce the concepts of normal and shear stress, and to use
them in the analysis and design of members subject to axial load
and direct shear.
•
7. 1
To define normal and shear strain, and show how they can be
determined for various types of problems.
INTRODUCTION
Mechanics of materials is a study of the internal effects of stress and
strain in a solid body that is subjected to an external loading. Stress is
associated with the strength of the material from which the body is made,
while strain is a measure of the deformation of the body. A thorough
understanding of the fundamentals of this subject is of vital importance
for the design of any machine or structure, because many of the formulas
and rules of design cited in engineering codes are based upon the
principles of this subject.
311
312
CHAPTER
7
S TR ES S AND STRAIN
7.2 INTERNAL RESULTANT LOADINGS
In order to design the horizontal
members of this building frame, it is first
necessary to find the internal loadings
at various points along their length.
When applying the methods of mechanics of materials, statics along with
the method of sections is used to determine the resultant loadings that act
within a body. For example, consider the body shown in Fig. 7- la, which
is held in equilibrium by the four external forces.* In order to obtain the
internal loadings acting on a specific region within the body, it is necessary
to pass an imaginary section or "cut" through the region where the
internal loadings are to be determined. The two parts of the body are then
separated, and a free-body diagram of one of the parts is drawn, Fig. 7- lb.
Here there is actually a distribution of internal force acting on the
"exposed" area of the section . These forces actually represent the effects
of the top section of the body acting on its bottom section.
Although the exact distribution of this internal loading may be
unknown, we can find its resultants FR and (MR)o at any specific point 0
on the sectioned area, Fig. 7- lc, by applying the equations of equilibrium
to the bottom part of the body. It will be shown later in the book that
point 0 is most often chosen at the centroid of the sectioned area, and so
we will always choose this location for 0 , unless otherwise stated. Also, if
a member is long and slender, as in the case of a rod or beam, the section
to be considered is generally taken perpendicular to the longitudinal axis
of the member. This section is referred to as the cross section.
F•
!
Section
(a)
(b)
(c)
Fi.g. 7-1
*111e body's weight is not shown, since it is assumed to be quite small, and therefore
negligible compared with the other loads.
7 .2
Torsional
moment
T
(MR)o ----- ------NNormal
force -- ,FR
Bending M ~.(:.}-;;:::~::-:;::::;:
moment
I
I
I
I
I
I
I
~--..~:v
Shear
force
(c)
(d)
Fig. 7-1 (cont.)
Three Dimensions. For later application of the formulas of mechanics
of materials, we will consider the components of FR and (MR)o acting
both normal and perpendicular to the sectioned area, Fig. 7- ld. Four
different types of resultant loadings can then be defined as follows:
Normal force, N. This force acts perpendicular to the area. It is
developed whenever the external loads tend to push or pull on the two
segments of the body.
Shear force, V. The shear force lies in the plane of the area and it is
developed when the external loads tend to cause the two segments of the
body to slide over one another.
Torsional moment or torque, T. This effect is developed! when
the external loads tend to twist one segment of the body with respect to
the other about an axis perpendicular to the area.
Bending moment, M. The bending moment is caused by the
external loads that tend to bend the body about an axis lying within the
plane of the area.
INTERNAL RESULTANT LOADINGS
31 3
314
CHAPTER
7
STRESS AND STRAIN
section
F,
(a)
(b)
Fi.g. 7- 2
Coplanar Loadings. If the body is subjected to a coplanar system of
forces, Fig. 7- 2a, then only normal-force, shear-force, and bending-moment
components will exist at the section, Fig. 7-2b. If we use the x, y, z
coordinate axes, as shown on the left segment, then N can be obtained
by applying IFr = 0, and V can be obtained from IFy = 0. Finally, the
bending moment M 0 can be determined by summing moments about
point 0 (the z axis), IM0 = 0, in order to eliminate the moments caused
by the unknowns N and V .
IMPORTANT POINTS
• Mechanics of materials is a study of the relationship between
the external loads applied to a body and the stress and strain
caused by the internal loads within the body.
• External forces can be applied to a body as distributed or
concentrated surface loadings, or as body forces that act
throughout the volume of the body.
• Linear distributed loadings produce a resultant force having a
magnitude equal to the area under the load diagram, and
having a location that passes through the centroid of this area.
• A support produces a force in a particular direction on its
attached member if it prevents translation of the member in
that direction, and it produces a couple moment on the member
if it prevents rotation.
• The equations of equilibrium I F = 0 and I M = 0 must be
satisfied in order to prevent a body from translating with
accelerated motion and from rotating.
• The method of sections is used to determine the internal
resultant loadings acting on the surface of a sectioned body. In
general, these resultants consist of a normal force, shear force,
torsional moment, and bending moment.
7 .2
PROCEDURE FOR ANALYSIS
The resultant internal loadings at a point located on the section of a
body can be obtained using the method of sections. This requires the
following steps.
Support Reactions.
• When the body is sectioned, decide which segment of the body
is to be considered. If the segment has a support or connection
to another body, then before the body is sectioned, it will be
necessary to determine the reactions acting on the chosen
segment. To do this, draw the free-body diagram of the entire
body and then apply the necessary equations of equilibrium to
obtain these reactions.
Free-Body Diagram.
• Keep all external distributed loadings, couple moments,
torques, and forces in their exact locations, before passing the
section through the body at the point where the resultant
internal loadings are to be determined.
• Draw a free-body diagram of one of the "cut" segments and
indicate the unknown resultants N, V, M, and T at the section.
These resultants are normally placed at the point representing
the geometric center or centroid of the sectioned area.
• If the member is subjected to a coplanar system of forces, only
N, V, and M act at the centroid.
• Establish the x, y, z coordinate axes with origin at the centroid
and show the resultant internal loadings acting along the axes.
Equations of Equilibrium.
• Moments should be summed at the section, about each of the
coordinate axes where the resultants act. Doing this eliminates the
unknown forces N and V and allows a direct solution for Mand T.
• If the solution of the equilibrium equations yields a negative
value for a resultant, the directional sense of the resultant JS
opposite to that shown on the free-body diagram.
The following examples illustrate this procedure numerically and also
provide a review of some of the important principles of statics.
INTERNAL RESULTANT LOADINGS
31 5
316
I
CHAPTER
EXAMPLE
7
STRESS AND STRAIN
7.1
Determine the resultant internal loadings acting on the cross section at C
of the cantilevered beam shown in Fig. 7- 3a.
270N/m
A
-
3 m ---t---- 6 m - - - (a)
Fig. 7-3
SOLUTION
The support reactions at A do not have to be
determined if segment CB is considered.
Support Reactions.
Free-Body Diagram. The free-body diagram of segment CB is
540N
Mc
Ne
~~~-- I
.-(:tic +-----------DI
180
I
Vcl-2m-I
(b)
4m
I
shown in Fig. 7- 3b. It is important to keep the distributed loading on
the segment until after the section is made. Only then should this
loading be replaced by a single resultant force. Notice that the intensity
of the distributed loading at C is found by proportion, i.e., from
Fig.7- 3a, w/6m = (270N/m)/9m,w = 180N/m.Themagnitudeof
the resultant of the distributed load is equal to the area under the
loading curve (triangle) and acts through the centroid of this area.
Thus, F = !(180 N/m)(6 m) = 540 N, which acts !(6 m) = 2 m from
C as shown in Fig. 7- 3b.
Equations of Equilibrium.
Applying the equations of equilibrium
we have
+jIF.y
=
C+IMc
=
O·,
O;
-Ne = 0
Ne = 0
Ve - 540N = 0
Ve = 540N
-Mc - 540N(2m) = 0
Mc = -1080N · m
Ans.
Ans.
Ans.
The negative sign indicates thlat M c acts in the opposite direction to
that shown on the free-body ,diagram. Try solving this problem using
segment AC, by first checking the support reactions at A, which are
given in Fig. 7- 3c.
7 .2
I
EXAMPLE
31 7
INTERNAL R ESULTANT LOADINGS
7.2
The 500-kg engine is suspended from the crane boom in Fig. 7-4a.
Determine the resultant internal loadings acting on the cross section of
the boom at point E.
r
D
1.5 m
SOLUTION
We will consider segment AE of the boom, so
we must first determine the pin reactions at A. Since member CD is a
two-force member, it acts like a cable, and therefore exerts a force Fco
having a known direction. The free-body diagram of the boom is
shown in Fig. 7-4b. Applying the equations of equilibrium,
Support Reactions.
Fco(~) (2 m) - [500(9.81) N](3 m)
Fco
=
=
l
0
(a)
12 262.5 N
Ax - (12 262.5N) (~)
=
0
Ax = 9810 N
A,
-A>'+ (12262.5 N)(~) - 500(9.81)N = 0
Ay
=
2452.5 N
500(9.81) N
Free-Body Diagram.
(b)
The free-body diagram of segment AE is
shown in Fig. 7-4c.
Equations of Equilibrium.
9810N
NE + 9810 N
NE = -9810 N
+ jIF.y
=
O·,
0
=
-9.81 kN
=
-VE - 2452.5 N
=
0
VE
=
ME
+ (2452.5N)(l m)
-2452.5 N
=
Ans.
-
lm -
2452.5 N
-2.45 kN
=
Ans.
Fig. 7-4
0
ME = -2452.5 N · m = -2.45 kN · m
(c)
Ans.
v£
318
I
CHAPTER
EXAMPLE
7
STRESS AND STRAIN
7.3
Determine the resultant internal loadings acting on the cross section at C
of the beam shown in Fig. 7- 5a.
9001b
i - - - - -8
300 lb/ft
ft
2
ft-1
----1
---
---
---
I
I
A,
(b)
(a)
Fig. 7-5
SOLUTION
Here we will consider segment BC, but first we
must find the force components at pin A. The free-body diagram of the
entire beam is shown in Fig. 7- 5b. Since member BD is a two-force
member, like member CD in Example 7.2, the force at B has a known
direction, Fig. 7- 5b. We have
Support Reactions.
~+IMA = O;
(900lb)(2ft) - (F8 vsin30°)10ft
=
0
F8 v
=
360lb
Using this result, the free-body diagram of
segment BC is shown in Fig. 7- 5c.
Free-Body Diagram.
1- ft-1
2
Mc
~.t;~c
Equations of Equilibrium.
~IF,
O;
=
=
0
Ne = 312lb
3601b
(c)
Ne - (360 lb) cos 30°
+f IFy
=
0;
Ans.
(360 lb) sin 30° - Ve = 0
Ve = 180 lb
~+IMc = O;
Mc - (360 lb) sin 30°(2 ft)
Mc
=
360 lb · ft
Ans.
=
O
Ans.
7 .2
I EX AMPLE
INTERNAL RESULTANT LOADINGS
7.4
Determine the resultant internal loadings acting on the cross section at B
of the pipe shown in Fig. 7-&1. End A is subjected to a vertical force of
50 N, a horizontal force of 30 N, and a couple moment of70 N · m. Neglect
the pipe's mass.
SOLUTION
The problem can be solved by considering segment AB, so we do not
need to calculate the support reactions at C.
Free·Body Diagram. The free-body diagram of segment AB is
shown in Fig. 7-6b, where the x, y, z axes are established at B. The
(a)
resultant force and moment components at the section are assumed
to act in the positive coordinate directions and to pass through the
centroid of the cross-sectional area at B.
Equations of Equilibrium.
Applying the six scalar equations of
equilibrium, we have*
IR..t = O·,
(Fs)x
= 0
Ans.
If',, = O;
(Fs)y + 30N = 0
(Fs)y = -30 N
Ans.
2.Fz = O;
(Fs)z - 50 N = 0
(Fs)z = SON
Ans.
(Ms)x + 70 N · m - 50 N (0.5 m) = 0
(Ms)x
2.(Ms)y = O;
=
-45 N · m
Ans.
(Ms) y + 50 N (1.25 m) = 0
(Ms)y = -62.5 N · m
(b)
Ans.
(Ms) z + (30 N)(l.25) = 0
(Ms)z = -37.5 N · m
Ans.
(Fs)Y' (Ms}" (Ms)y, and (Ms)z
indicate? The normal force Ns = I(Fs) vI = 30 N, whereas the shear
force is Vs = V(0) 2 + (50) 2 = 50 N. Also, the torsional moment is
Ts = I(Ms) yl = 62.5 N · m, and the bending moment is Ms =
V (45)2 + (37.5)2 = 58.6 N · m.
NOTE: What do the negative signs for
*The magnimde of each moment about the x, y , or z axis is equal to the magnitude
of each force times the perpendicular distance from the axis to the line of action of
the force. The direction of each moment is detem1ined using the right-hand rule, with
positive moments (thumb) directed along the positive coordinate axes.
Fig. 7-6
31 9
320
C HAPT ER
7
S TR ESS AND STRAIN
PRELIMINARY PROBLEM
P7-1. In each case, explain how to find the resultant internal
loading acting on the cross section at point A. Draw all
necessary free-body diagrams, and indicate the relevant
equations of equilibrium. Do not calculate values. The lettered
dimensions, angles, and loads are assumed to be known.
D
0
p
(d)
A
f!--:------l!f------T---~1 c
1---20---1--a-l-a-I
c
p
(a)
IV
•
c
A
a
a
(e)
p
a
B
-
/"
'
c
(b)
I
3a
-r
a
IA •
l
-I
a
I
D
I
81., •P.-i~a~-1
--·l
C
a
_j ~
p
1o--a-I B
'
p
(c)
(f)
Prob.P7-1
-
7.2
3 21
INTERNAL R ESULTANT LOADINGS
FUNDAMENTAL PROBLEMS
.'7-1. Determine the internal normal force, shear force,
and bending moment at point C in the beam.
17-4. Determine the internal normal force, shear force,
and bending moment at point C in the beam.
lO k N/m
IOkN
60kN·m
A
,
1c
i - - -3
-
B
m----3 m-----1
Prob. •'7-4
Prob. l 7-1
F7-5. Determine the internal normal force, shear force,
and bending moment al point C in the beam.
F7-2. Determine the internal normal force, shear force,
and bending moment at point C in the beam.
300 lb/ft
OON/
A
200N/m
kl iI (111JI!111111IB
_JC
•
I
~
l-1.sm
i>r..... I /-.,,
1.sm-l
Prub.
1-2
Determine the internal normal force, shear force,
and bending moment at point C in the beam.
.'7-6.
17-3. Determine the internal normal force, shear force,
and bending moment at point C in the beam.
20 kN/m
.......
~2 m~l-2 m-l-2 m-l
Prob. 17-3
3m
_L lf]):o=~
1° 2m-l -2 m=-l- 2 m Prob. F7-6
322
CHAPTER
7
S TR ES S AND STRAIN
PROBLEMS
7-1. The shaft is supported by a smooth thrust bearing
at B and a journal bearing at C. Determine the resultant
internal loadings acting on the cross section at £.
*7-4. The shaft is supported by a smooth thrust bearing
at A and a smooth journal bearing at B. Determine the
resultant internal loadings acting on the cross section at C.
600N/m
E
C
B
c
4ft j -4ft -
lm
4001b
-Lsml sm
8001b
900N
Prob. 7-1
7-2. Determine the resultant internal normal and shear
force in the member at (a) section a-a and (b) section b-b,
each of which passes through point A. The 500-lb load is
applied along the centroidal axis of the member.
a
Prob. 7-4
7-5. Determine the resultant internal loadings acting on
the cross section at point B.
60 lb/ft
b
300
A
SOOJb . - - -1-
-+---.SOOlb
- + - - - - - - 12 ft - - - - - A
a
Prob. 7-5
Prob. 7-2
7-6. Determine the resultant internal loadings on the
cross section at point D.
7-3. Determine the resultant internal loadings acting on
section b-b through the centroid, point Con the beam.
7-7. Determine the resultant internal loadings at cross
sections at points E and Fon the assembly.
1 - -1.S m - - - 1
Prob. 7-3
Probs. 7-617
7.2
*7-8. The beam supports the distributed load shown.
Determine the resultant internal loadings acting on the cross
section at point C. Assume the reactions at the supports A
and B are vertical.
323
INTERNAL RESULTANT LOADINGS
7-11. Determine the resultant internal loadings acting on
the cross sections at points D and £of the frame.
*7-ll. Determine the resultant internal loadings acting
on the cross sections at points F and G of the frame.
7-9. The beam supports the distributed load shown.
Determine the resultant internal loadings acting on the cross
section at point D. Assume the reactions at the supports A
and B are vertical.
T
4 ft
l
4kN/m
A ~::=::==-:~:...:=::=::=::=::=~::=:::::;;~ B
1-
t.5 m-·1--C-- 3 m
_J_~5
m
-1
150lb
Probs. 7-11112
Probs. 7-S/9
7-10. The boom DF of the jib crane and the column DE
have a uniform weight of SO lb/ft.Uthe hoist and load weigh
300 lb, determine the resultant internal loadings in the crane
on cross sections at points A. 8. and C.
7-13. The blade of the hacksaw is subjected to a pretension
force of F= 100 N. Determine the resultant internal loadings
acting on section a-a at point D.
7-14. The blade oft he hacksaw is subjected to a pretension
force of F= 100 N. Determine the resultant internal loadings
acting on section b-b at point D.
-.---J2.D',.E:::';'j== = ===;=.Ar= F
a
5 ft
-225mm300 b
-..
-+----+f
300 lb
7 ft
E
Prob. 7-10
Probs. 7-13/14
324
CHAPTER
7
STRESS ANO STRAIN
7-15. The beam supports the triangular distributed load
shown. Determine the resultant internal loadings on the
cross section at point C. Assume lhe reactions at the supports
A and B are vertical.
*7-16. The beam supports the dislributed load shown.
Determine the resultant internal loadings on the cross
sections at points D and £. Assume the reactions at the
supports A and B are vertical.
7-18. The shaft is suppo rt ed at ils ends by two bearings
A and Band is subjected to the forces applied to lhe pulleys
fixed to the shaft. Determine the resultant internal loadings
acting on the cross section at point C. The 400-N forces act
in the - z dir ection and the 200-N and 80-N forces act in the
+y direction. The journal bearings at A and B exert only y
and z components of force on the shaf1.
800 lb/ft
y
C
- -- 1 - - -1- 6
6 ft
6 fl
E
n-l-4.5-·1
- --1
ft 4.5 ft
Probs. 7- 15/16
x
Prob. 7-18
7-17. The shaft is supported at its ends by two bearings
A and B and is subjected to lhe forces applied to the pulleys
fixed to the shaft. Determine the resultant internal loadings
acting on the cross section at point D. The 400-N forces act in
the - z direction and the 200-N and 80-N forces act in the +y
direction. The journal bearings at A and B exert only y and z
components of force on the shaf1.
7-19. The hand crank that is used in a press bas the
dimensions shown. Determine the resultant internal
loadings acting on the cross section at point A if a vertical
force of 50 lb is applied to the handle as shown. Assume the
crank is fLxed to the shaft at B.
8
y
y
x
501b
Prob. 7-17
Prob. 7-19
7 .2
*7- 20. Determine the resultant internal loadings acting on
the cross section at point C in the beam. The load D has a
mass of 300 kg and is being hoisted by the motor M with
constant velocity.
INTERNAL R ESULTANT LOADINGS
325
*7- 24. Determine the resultant internal loadings acting
on the cross section at point C. The cooling unit has a total
weight of 52 kip and a center of gravity at G.
7- 2L Determine the resultant internal loadings acting on
the cross section at point E. The load D has a mass of 300 kg
and is being hoisted by the motor M with constant velocity.
F
1 -201 - J -201 - 1 -201 - J
0.101
0.101
I
I
•
C
E
L 101 -i-
A
1.501 -
D
c
A
II--- -
B
3 ft ---~---- 3 ft -------
Probs. 7- 20/21
•
7- 22. The metal stud punch is subjected to a force of 120 Non
the handle. Determine the magnitude of the reactive force at
the pin A and in the short link BC. Also, determine the resultant
internal loadings acting on the cross section at point D.
7- 23. Determine the resultant internal loadings acting on the
cross section at point E of the handle arm, and on the cross
section of the short link BC.
Prob. 7-24
7- 25. Determine the resultant internal loadings acting on
the cross section at points B and C of the curved member.
120N
A
500 Jb
Probs. 7- 22123
Prob. 7-25
326
CHAPTER
7
STRESS AND STRAIN
7.3
STRESS
It was stated in Section 7.2 that the force and moment acting at a specified
point 0 on the sectioned area of the body, Fig. 7- 7, represents the
resultant effects of the distribution of loading that acts over the
sectioned area, Fig. 7- 8a. Obtaining this distribution is of primary
importance in mechanics of materials. To solve this problem it is first necessary
to establish the concept of stress.
We begin by considering the sectioned area to be subdivided into
small areas, such as dA shown in Fig. 7- 8a. As we reduce dA to a smaller
and smaller size, we will make two assumptions regarding the
properties of the material. We will consider the material to be
continuous, that is, to consist of a continuum or uniform distribution of
matter having no voids. Also, the material must be cohesive, meaning
that all portions of it are connected together, without having breaks,
cracks, or separations. A typical finite yet very small force dF, acting on
dA , is shown in Fig. 7-8a. This force, like all the others, will have a
unique direction, but to compare it with all the other forces, we will replace
it by its three components, namely, d F., d F>" and d Fz. As ~A approaches
zero, so do dF and its components; however, the quotient of the force and
area will approach a finite limit. This quotient is called stress, and
it describes the intensity of the internal force acting on a specific plane
(area) passing through a point.
Fig. 7-7
z
I
',
'', 6F
6F,
'''
:'
6F
\ /
~­
\
y
x
(a)
y
x
(b)
Fig. 7-8
y
x
(c)
7.3
STRESS
327
Normal Stress. The intensity of the force acting normal to ~A is
referred to as the normal stress, a (sigma). Since dFz is normal to the
area then
(7- 1)
If the normal force or stress "pulls" on dA as shown in Fig. 7-8a, it is
tensile stress, whereas if it "pushes" on dA it is compressive stress.
Shear Stress. The intensity of force acting tangent to ~A is called the
shear stress, -r (tau). Here we have two shear stress components,
z
'T
zx
=
I
.
dfr
Jim - -
u,
aA~OdA
(7-2)
'Tzy
The subscript notation z specifies the orientation of the area aA,
Fig. 7- 9, and x and y indicate the axes along which each shear stress acts.
----y
Fig.7-9
General State of Stress. If the body is further sectioned by planes
parallel to the x- z plane, Fig. 7-8b, and the y- z plane, Fig. 7-8c, we can
then "cut out" a cubic volume element of material that represents the
state of stress acting around a chosen point in the body. This state of
stress is then characterized by three components acting on each face of
the element, Fig. 7- 10.
Units. Since stress represents a force per unit area, in the International
Standard or SI system, the magnitudes of both normal and shear stress
are specified in the base units of newtons per square meter (N/m2 ) . This
combination of units is called a pascal (1 Pa = 1 N /m2 ), and because it is
rather small, prefixes such as kilo- (103) , symbolized by k, mega- (106) ,
symbolized by M, or giga- (109), symbolized by G, are used in engin eering
to represent larger, more realistic values of stress.* In the Foot-PoundSecond system of units, engineers usually express stress in pounds per
square inch (psi) or kilopounds per square inch (ksi), where 1 kilopound
(kip) = 1000 lb.
*Sometimes stress is expressed in units of N/mm2 , where 1 mm = 10- 3 m. However, in
the SJ system, prefixes are not allowed in the denominator of a fraction, and therefore it
is better to use the equivalent 1 N/mm2 = 1 MN/m2 = 1 MPa.
y
Fig. 7-10
328
C HAPT ER
7
S TR ESS AND STRAIN
7.4
AVERAGE NORMAL STRESS IN AN
AXIALLY LOADED BAR
We will now determine the average stress distribution acting over the
cross-sectional area of an axially loaded bar such as the one shown in
Fig. 7- lla. Specifically, the cross section is the section taken perpendicular
to the longitudinal axis of the bar, and since the bar is prismatic all cross
sections are the same throughout its length. Provided the material of the
bar is both homogeneous and isotropic, that is, it has the same physical
and mechanical properties throughout its volume, and it has the same
properties in all directions, then when the load P is applied to the bar
through the centroid of its cross-sectional area, the bar will deform
uniformly throughout the central region of its length, Fig. 7- llb .
Realize that many engineering materials may be approximated as
being both homogeneous and isotropic. Steel, for example, contains
thousands of randomly oriented crystals in each cubic millimeter of its
volume, and since most objects made of this material have a physical size
that is very much larger than a single crystal, the above assumption
regarding the material's composition is quite realistic.
Note that anisotropic materials, such as wood, have different properties
in different directions; and although this is the case, if the grains of wood
are oriented along the bar's axis (as for instance in a typical wood board),
then the bar will also deform uniformly when subjected to the axial load P.
Average Normal Stress Distribution. If we pass a section
through the bar, and separate it into two parts, then equilibrium requires
the resultant normal force Nat the section to be equal to P, Fig. 7- llc.
And because the material undergoes a uniform deformation, it is necessary
that the cross section be subjected to a constant normal stress distribution.
p
p
~
,t_
Region of
uniform
deformation
of bar
N =P
Cross-sectional
area
i
.._ ....
i
p
i
p
p
(a)
(b)
(c)
Fig. 7-11
External force
7.4
AVERAGE NORMAL STRESS IN AN AXIAUY LOADED
As a result, each smaU area 6.A on the cross section is subjected to a
force 6.N = u 6.A , Fig. 7-lld, and the sum of these forces acting over
the entire cross-sectional area must be equivalent to the internal resultant
force P at the section. U we let 6.A ~ dA and therefore 6.N ~ dN, then,
recognizing u is constant, we have
N
BAR
z
I
N
= CTA
y
(7-3)
x
H ere
CT=
average no rma l stress at any point on the cross-sectional area
N =internal resultant normal force, which acts through the centroid of the
cross-sectional area. N is determined using the method of sections
and the equations of equilibrium, where for this case N = P.
A =cross-sectional area of the bar where CT is determined
Equilibrium The stress distribution in Fig. 7- 11 indicates that only a
normal stress exists on any small volume element of material located at
each point on the cross section. Thus, if we consider vertical equilibrium
of an element of material and then apply the equation of force
equilibrium to its free-body diagram, Fig. 7-12,
'i.F.z
= O·,
CT(M) - CT'(M)
u
=
=
0
u'
uM
u
f
r
L
CT
Stress on element
Free-body diagram
Fig. 7-U
(d)
Fig. 7-11 (cont.)
329
330
CHAPTER
7
STRESS AND STRAIN
N
N
t
U=
i
N
A
a
qi
t
t
i
p
p
Tension
Compression
Fig. 7-13
In other words, the normal stress components on the element must be
equal in magnitude but opposite in direction. Under this condition the
material is subjected to uniaxial stress, and this analysis applies to
members subjected to either tension or compression, as shown in
Fig. 7- 13.
Although we have developed this analysis for prismatic bars, this
assumption can be relaxed somewhat to include bars that have a slight
taper. For example, it can be shown, using the more exact analysis of the
theory of elasticity, that for a tapered bar of rectangular cross section,
where the angle between two adjacent sides is 15°, the average normal
stress, as calculated by u = N /A, is only 2.2°/o less than its value found
from the theory of elasticity.
This steel tie rod is used as a hanger to
suspend a portion of a staircase, and as
a result it is subjected to tensile stress.
Maximum Average Normal Stress. For our analysis, both the
internal force N and the cross-sectional area A were constant along the
longitudinal axis of the bar, and as a result the normal stress u = N /A is
also constllnt throughout the bar's length. Occasionally, however, the bar
may be subjected to several external axial loads, or a change in its crosssectional area may occur. As a result, the normal stress within the bar
may be different from one section to the next, and, if the maximum
average normal stress is to be determined, then it becomes important to
find the location where the ratio N /A is a maximum. Example 7.5
illustrates the procedure. Once the internal loading throughout the bar is
known, the maximum ratio N /A can then be identified.
7.4
A VERAGE N ORMAL STRESS IN AN AXIAUY LOADED
IMPORTANT POINTS
• When a body subjected to external loads is sectioned, there is a
distribution of force acting over the sectioned area which holds
each segment of the body in equilibrium. The intensity of this
internal force at a point in the body is referred to as stress.
• Stress is the limiting value of force per unit area, as the area
approaches zero. For this definition, the material is considered to
be continuous and cohesive.
• The magnitude of the stress components at a point depends upon
the type of loading acting on the body, and the orientation of the
element at the point.
• When a prismatic bar is made of homogeneous and isotropic
material, and is subjected to an axial force acting through the
centroid of the cross-sectional area, then the center region of the
bar will deform uniformly. As a result, the material will be
subjected only to normal stress. This stress is uniform or averaged
over the cross-sectional area.
PROCEDURE FOR ANALYSIS
The equation <T = N /A gives the average normal stress on the
cross-sectional area of a member when the section is subjected to
an internal resultant normal force N. Application of this equation
requires the following steps.
Internal Loading.
• Section the member perpendicular to its longitudinal axis at
the point where the normal stress is to be determined, and
draw the free-body diagram of one of the segments. Apply the
force equation of equilibrium to obtain the internal axial force
N at the section.
A verage Normal Stress.
• Determine the member's cross-sectional area at the section
and calculate the average normal stress u = N / A.
• It is suggested that u be shown acting on a small volume element
of the material located at a point on the section where stress is
calculated. To do this, first draw u on the face of the element
coincident with the sectioned area A. Here u acts in the same
direction as the internal force N since all the normal stresses on
the cross section develop this resultant. The normal stress u on
the opposite face of the element acts in the opposite direction.
BAR
3 31
332
CHAPTER
EXAMPLE
7
7.5
~
STRESS AND STRAIN
-
The bar in Fig. 7- 14a has a constant width of 35 mm and a thickness of
10 mm. Determine the maximum average normal stress in the bar when
it is subjected to the loading shown.
_,,
12kN
A
_L
B
c
9kN
4 kN
D
22kN
~
35mm
9kN
4 kN
(a)
12 kN II
~
,QJ
.i
.. NAB= 12kN
9kN
~
12kN
N 8 c = 30kN
9kN
N(kN)
9kN
12 kN II
30
22
12
~I
O!iN
x
4 kN
~
Nc0 =22kN
4kN
(b)
(c)
SOLUTION
Internal Loading. By inspection, the internal axial forces in regions
AB, BC, and CD are all constant yet have different magnitudes. Using
the method of sections, these loadings are shown on the free-body
diagrams of the left segments shown in Fig. 7- 14b. *The normal force
diagram , which represents these results graphically, is shown in
Fig. 7- 14c. The largest loading is in region BC, where N 8 c = 30 kN.
Since the cross-sectional area of the bar is constant, the largest average
normal stress also occurs within this region of the bar.
Average Normal Stress. Applying Eq. 7- 3, we have
30(HY) N
(0.035 m)(0.010 m)
35mm
5.7 MPa
(d)
=
85 7
· MPa
Ans.
The stress distribution acting on an arbitrary cross section of the bar
within region BC is shown in Fig. 7- 14d.
Fig. 7-14
*Show that you get these same results using the right segments.
7.4
I
EXAMPLE
AVERAGE NORMAL STRESS IN AN AxlALLY LOADED BAR
7.6
The 80-kg lamp is supported by two rods AB and BC as shown in
Fig. 7- 15a. If AB has a diameter of 10 mm and BC has a diameter of
8 mm, determine the average normal stress in each rod.
y
Fsc
(a)
80(9.81) = 784.8 N
(b)
Fig. 7-15
SOLUTION
We must first determine the axial force in each rod.
A free-body diagram of the lamp is shown in Fig. 7- 15b. Applying the
equations of force equilibrium,
Internal Loading.
~ 2F, = O;
F8 c( ~)
+ f2Fy
F8 c( ~) + F8 Asin60° - 784.8N = 0
=
O;
-
F8A cos 60° = 0
F8 c = 395.2 N,
FBA = 632.4 N
By Newton's third Jaw of action, equal but opposite reaction, these
forces subject the rods to tension throughout their length.
Average Normal Stress.
Applying Eq. 7- 3,
8.05 MPa
asc
=
_Fs_c
=
Ase
FBA
aBA = =
AsA
395.2 N
?T(0.004 m) 2
632.4 N
)2
1T 0.005 m
(
=
7.86 MPa
Ans.
=
8.05MPa
Ans.
The average normal stress distribution acting over a cross section of
rod AB is shown in Fig. 7- 15c, and at a point on this cross section, an
element of material is stressed as shown in Fig. 7- 15d.
8.05 MPa
(d)
333
334
CHAPTER
EXAMPLE
-
7
7.7
STRESS AND STRAIN
Member AC shown in Fig. 7- 16ll. is subjected to a vertical force of 3 kN.
Determine the position x of this force so that the average compressive
stress at the smooth support C is equal to the average tensile stress in the
tie rod AB. The rod has a cross-sectional area of 400 mm2 and the contact
area at C is 650 mm2 .
xA
c
(b)
(a)
Fi.g. 7-16
SOLUTION
Internal Loading. The forces at A and C can be related by considering
the free-body diagram of member AC, Fig. 7- 16b. There are three unknowns,
namely, FA 8 , Fe, and x. To solve we will work in units of newtons and
millimeters.
+f 2£,, =
FAB + Fe - 3000 N
-3000 N(x) + Fc(200 mm)
0;
C+2MA = O;
=
=
0
0
(1)
(2)
Average Normal Stress. A necessary third equation can be written
that requires the tensile stress in the bar AB and the compressive stress
at C to be equivalent, i.e.,
a =
FAB
2
400 mm
Fe
=
=
Fe
650 mm2
1.625FA 8
Substituting this into Eq. 1, solving for FAB , then solving for Fe, we obtain
FAB = 1143 N
Fe = 1857N
The position of the applied load is determined from Eq. 2,
x = 124mm
As required, 0 < x < 200 mm.
Ans.
7.5
AVERAGE SH EAR STRESS
7.5 AVERAGE SHEAR STRESS
335
F
!
Shear stress has been defined in Section 7.3 as the stress component that
acts in the plane of the sectioned area. To show how this stress can develop,
consider the effect of applying a force F to the bar in Fig. 7- 17a. If F is
Large enough, it can cause the material of the bar to deform and fail along
the planes identified by A 8 and CD. A free-body diagram of the
unsupported center segment of the bar, Fig. 7-17b, indicates that the shear
force V = F/2 must be applied at each section to hold the segment in
equilibrium. The average shear stress distributed over each sectioned area
that develops this shear force is defined by
c
(a)
F
i
[
I
v
(7--4)
v
(b)
F
h
1 t~Tavg
I I
I I
Here
= average shear stress at the section, which is assumed to be the
same at each point on the section
V =internal resultant shear force on the section determined from
the equations of equilibrium
A =area of the section
Tavg
The distribution of average shear stress acting over the sections is
shown in Fig. 7-17c. Notice that T3 ,.8 is in the same direction as V, since
the shear stress must create associated forces, all of which contribute to
the internal resultant force V.
The loading case discussed he re is an example of simple or direct
shear, since the shear is caused by the direct action of the applied load F.
This type of shear often occurs in various types of simple connections
that use bolts, pins, welding material, etc. In all these cases, however,
application of Eq. 7--4 is only approximate. A more precise investigation
of the shear-stress distribution over the section often reveals that
much larger shear stresses occur in the material than those predicted
by this equation. A lthough this may be the case, application of Eq. 7--4
is generally acceptable for many problems involving the design or
analysis of small e le me nts. For example, engineering codes allow its
use for determining the size or cross section of fasteners such as bolts,
and for obtaining the bonding strength of glued joints subjected to
shear loadings.
(c)
Fig. 7-17
The pin A used to connect the
linkage of this tractor is subjected to
double sh ear because s hea ring
stresses occur on the surface of the
pin at 8 and C. See Fig. 7-19c.
336
7
CHAPTER
STRESS AND STRAIN
z
f
z
,.-·-----~
A>,
(r
Section plane
_ _,.
7 :.y
11.x
-----v
0
l
11.z
Section plane
11.y
uX
A
r,
;
-·~-)'
•.., , , _ _
•
'T.;:y
Free-body diagram
(b)
(a)
(c)
Fig. 7-18
Shear Stress Equilibrium.
Let us consider the block in
Fig. 7- 18a, which has been sectioned and is subjected to the internal
shear force V. A volume element taken at a point located on its surface
will be subjected to a direct shear stress 7:ry> as shown in Fig. 7- 18b.
However, force and moment equilibrium of this element will also require
shear stress to be developed on three other sides of the element. To show
this, it is first necessary to draw the free-body diagram of the element,
Fig. 7- 18c. Then force equilibrium in they direction requires
rorce
1
1
stress area
~F.y =
O·,
n1
I
'Tzy (~x~y) - 'T~y ~x~y = 0
'Tzy = 'T~y
In a similar manner, force equilibrium in the
Fmally, taking moments about the x axis,
--·
z direction yields 'Tyz
= T~z·
moment
'T
I
force
I
amt
s~rea L-,
111
~Mx =
O·,
·1
n
-Tzy (~x ~y) ~z
+
'Tyz (~x ~z) ~y = 0
'T
In other words,
Pure shear
(d)
and so, all four shear stresses must have equal m agnitude and be directed
either toward or away from each other at opposite edges of the element,
Fig. 7- 18d. This is referred to as the complementary property of shear,
and the element in this case is subjected to pure shear.
7.5
IMPORTANT POINTS
• If two parts are thin or small when joined together, the applied
loads may cause shearing of the material with negligible
bending. If this is the case, it is generally assumed that an
average shear stress acts over the cross-sectional area.
• When shear stress T acts on a plane, then equilibrium of a
volume element of material at a point on the plane requires
associated shear stress of the same magnitude act on the three
other sides of the element.
PROCEDURE FOR ANALYSIS
The equation Tavg = V /A is used to determine the average shear
stress in the material. Application requires the following steps.
Internal Shear.
• Section the member at the point where the average shear stress
is to be determined.
• Draw the necessary free-body diagram, and calculate the
internal shear force V acting at the section that is necessary to
hold the part in equilibrium.
Average Shear Stress.
• Determine the sectioned area A , and then calculate the
average shear stress Tavg = V / A.
• It is suggested that Tavg be shown on a small volume element of
material located at a point on the section where it is determined.
To do this, first draw Tavg on the face of the element, coincident
with the sectioned area A. This stress acts in the same direction
as V.The shear stresses acting on the three adjacent planes can
then be drawn in their appropriate directions following the
scheme shown in Fig. 7-18d.
AVERAGE SHEAR STRESS
337
338
CHAPTER
EXAMPLE
-
7
7.8
S TR ES S AND STRAIN
Determine the average shear stress in the 20-mm-diameter pin at A and
the 30-mm-diameter pin at B that support the beam in Fig. 7- 19a.
SOLUTION
Internal Loadings. The forces on the pins can be obtained by
considering the equilibrium of the beam, Fig. 7- 19b.
C+IMA
•
A
~
A,
t 30kN5
y
AB
~4
!
9
~2m -l-4m -1
(b)
=
O;
F~~)(6m)
~If, = O;
(12.5
+ fIFy
A>'+ (12.5
=
O;
- 30kN(2m)
=
Fs
0
kN)(~) -Ax = 0
kN)(~) -
30 kN
=
12.5 kN
Ax = 7.50 kN
=
0
A>' = 20 kN
Thus, the resultant force acting on pin A is
FA
0
(c)
=
VA;+ A~ = V(7.50 kN) 2 + (20 kN) 2
=
21.36 kN
The pin at A is supported by two fixed "leaves" and so the free-body
diagram of the center segment of the pin shown in Fig. 7- 19c has two
shearing surfaces between the beam and each leaf. Since the force of
the beam (21.36 kN) acting on the pin is supported by shear force on
each of two surfaces, it is called double shear. Thus,
VA = FA = 21.36 kN = 10.68 kN
2
2
In Fig. 7- 19a, note that pin Bis subjected to single shear, which occurs
on the section between the cable and beam, Fig. 7- 19d. For this pin
segment,
Vs
=
Fs
12.5 kN
=
Average Shear Stress.
10.68(103 ) N
VA
(TA)avg
(d)
Fig. 7-19
= A
=
=
A
1T(002m) 2
Vs
12.5(103 ) N
(Ts ).vg = - =
As
4
1T (
4
34.0 MPa
Ans.
.
0 03 m) 2
.
= 17.7 MPa
Ans.
7.5
EXAMPLE
AVERAGE SHEAR STRESS
7 .9
If the wood joint in Fig. 7-20a has a thickness of 150 mm, determine the
average shear stress along shear planes a-a and b-b of the connected
member. For each plane. represent the state of stress on an element of
the material.
--
a
tD!
a
'•=200kPa -
0.1 m
iDt
b
b
0.125 m
I
(c)
3kN
-
-r6 = 160kPa ~
(a)
(d)
6 kN
F
(b)
Fi.g. 7-20
SOLUTION
Referring to the free-body diagram of the member,
Fig. 7-20b,
±. lF,.
= O;
6kN-F-F=O
F = 3kN
Now consider the equilibrium of segments cut across shear planes a-a
and b-b, shown in Figs. 7-20c and 7- 20d.
±.If,= O;
Va-3kN=O
= O·,
3 kN - Vb= 0
+
~
-~F..r
Average Shear Stress.
Yi,
(Ta) avg= Aa
=
3(HP) N
(0.1 m) (0.15 m)
=
200 kPa
Ans.
3( 103 ) N
('Tb) avg =
=- - - - --- = 160 kPa
Ans.
Ji
(0.125 m) (0.15 m)
The state of stress on elements located on sections a-a and b-b is shown
in Figs. 7-20c and 7-20d, respectively.
Vb
-A
•
[o
-+--
vb
F
Internal Loadings.
v.
6kN
6kN
JkN
]
339
340
CHAPTER
7
S TR ES S AND STRAIN
PRELIMINARY PROBLEMS
P7-2. In each case, determine the largest internal shear
force resisted by the bolt. Include all necessary free-body
diagrams.
P7-4. Determine the internal normal force at section A if
the rod is subjected to the external uniformally distributed
loading along its length of 8 kN / m.
, ( 8 kN/m
A
1- 2m
1-- - 3 m _ _ __,
Prob. P7-4
P7- S. The lever is held to the fixed shaft using the pin AB.
If the couple is applied to the lever, determine the shear
force in the pin between the pin and the lever.
(a)
B
6kN
lOkN
4kN
8kN
20kN
!
0.2m~
t
0.2 m
20N
20N
Prob. P7- S
(b)
Prob. P7-2
P7-6. The single-V butt joint transmits the force of 5 kN
from one bar to the other. Determine the resultant normal and
shear force components on the face of the weld, section AB.
P7- 3. Determine the largest internal normal force in the bar.
SkN
120mm
F
c
D
I
I
5 kN
B
I
A
I
2kN
6kN
Prob. P7- 3
I•
lO kN
v
Prob. P7-6
~
lOOmm
5 kN
7.5
341
AVERAGE SHEAR STRESS
FUNDAMENTAL PROBLEMS
F7-7. The uniform beam is supported by two rods AB and
CD that have cross-sectional areas of 10 mm2 and 15 mm2 ,
respectively. Determine the intensity w of the distributed
load so that the average normal stress in each rod does not
exceed 300 kPa.
F7-10. If the 600-kN force acts through the centroid of
the cross section, determine the location y of the centroid
and the average normal stress on the cross section. Also,
sketch the normal stress distribution over the cross section.
600kN
B
D
JV
·6 m _ _ _ _ ____,
A l·~----
C
Prob. F7-7
Prob. F7-10
F7-8. Determine the average normal stress on the cross
section. Sketch the normal stress distribution over the
cross section.
300 kN
F7-1L Determine the average normal stress at points A, B,
and C. The diameter of each segment is indicated in the figure.
•
3 kip
I
~
0.5 in.
1 in.
I
0.5 in.
I___»_._ ......I: ~
: 9 ki; ...
kip :
I
)I
2kip
lOOmm
Prob. F7-11
F7-12. Determine the average normal stress in rod AB if the
load has a mass of 50 kg. The diameter of rod AB is 8 mm.
Prob. F7-8
F7-9. Determine the average normal stress on the cross
section. Sketch the normal stress distribution over the
cross section.
15 kip
/?~
4 in.
1 in.
4 in. ~
--....,...::...1:.;in.
77·
1 in.
B
0
I
8mm
~
Prob.F7-9
Prob. F7-12
342
CHAPTER
7
S TR ES S AND STRAIN
PROBLEMS
7- 26. The supporting wheel on a scaffold is held in place
on the leg using a 4-mm-diameter pin. If the wheel is
subjected to a normal force of 3 kN, determine the average
shear stress in the pin. Assume the pin only supports the
vertical 3-kN load.
*7- 28. The bar has a cross-sectional area A and is
subjected to the axial load P. Determine the average normal
and average shear stresses acting over the shaded section,
which is oriented at (J from the horizontal. Plot the variation
of these stresses as a function of 9 ( 0 < 9 < 90°).
Prob. 7-28
7-29. The small block has a thickness of 0.5 in. If the stress
distribution at the support developed by the load varies as
shown, determine the force F applied to the block, and the
distanced to where it is applied.
F
__..d--\
3kN
\
Prob. 7- 26
7- 27. Determine the largest intensity w of the uniform
loading that can be applied to the frame without causing
either the average normal stress or the average shear stress
at section b-b to exceed u = 15 MPa and -r = 16 MPa,
respectively. Member CB has a square cross section of
30 mm on each side.
B
1.5 in.
y-
Prob. 7-29
7- 30. If the material fails when the average normal stress
reaches 120 psi, determine the largest centrally applied
vertical load P the block can support.
7- 31. If the block is subjected to a centrally applied force of
P = 6 kip, determine the average normal stress in the
material. Show the stress acting on a differential volume
element of the material.
1.
4 in.~
1
Prob. 7- 27
.
uzs
~lin.
!
P
.
12m.
~in.
Probs. 7- 30/31
7.5
*7-32. The plate has a widtJ1 of0.5 m. Uthe stress distribution
at the support varies as shown, determine the force P applied
to the plate and the distanced lo where it is applied.
:m=n
"l_
343
AVERAGE SHEAR STRESS
7- 35. Determine the average normal stress in each of the
20-mm-diameter bars of the truss. Set P =40 kN.
*7-36. If the average normal stress in each of the
20-mm-<iiameter bars is not allowed to exceed 150 MPa.
determine the maximum force P that can be applied to joint C.
7-37. Determine the maximum average shear stress in pin
A of the truss. A horizontal force of P = 40 kN is applied to
joint C. Each pin has a diameter of 25 mm and is subjected
to double shear.
~
-
u - (15.r Ifl) MPa_.../
30MPa
c
Prob. 7-32
7-33. The board is subjected to a tensile force of 200 lb.
Determine the average normal and average shear stress in
the wood fibers, which arc oriented along plane a-a at 20°
with the axis of the board.
1- - - - - 2 m Probs. 7-35136137
Prob. 7-33
7-34. The boom has a uniform weight of 600 lb and is
hoisted into position using the cable BC. Uthe cable has a
diameter of O.S in.. plot the average normal stress in the
cable as a function of the boom position 8 for 0° < 8 < 90°.
----1
7- 38. If P = S kN, determine the average shear stress in
the pins at A, B, and C. All pins are in double shear, and
each has a diameter of 18 mm.
7-39. Determine the maximum magnitude P of the loads
the beam can support if the average shear stress in each pin
is not to exceed 80 MPa. All pins arc in double shear. and
each has a diameter of 18 mm.
6P
p
3P
p
0.5 m
0.5 m
~
--t.5m--2m--l.5m - -1
8
A
c
Prob. 7-34
Probs. 7-38139
344
CHAPTER
7
STRESS AND STRAIN
*7-40. The column is made of concrete having a density
of 2.30 Mg/ m3. At its top B it is subjected to an axial
compressive force of 15 kN. Determine the average normal
stress in the column as a function of the distance z measured
from its base.
7-43. UP= 15 kN, determine the average shear stress in
the pins at A. B. and C. All pins are in double shear. and
each has a diameter of 18 mm.
T
+
P
4P
4P
2P
0.Sm
I
0.5m
- -1 m-+-t.5 m
1.5 m-
B
'
15kN
Prob. 7-43
*7-44. The railcar docklight is supported by the tin-diameter
pin at A. If the lamp weighs 4 lb, and the extension arm A 8 has
a weight of 0.5 lb/ft, determine the average shear stress in the
pin needed to support the lamp. Hint: The shear force in the pin
is caused by the couple moment required for equilibrium at A.
4m
z
- - - -- - - 3 ft - - -- -- - 1
1
A
x
}'
Prob. 7-40
1.25 in.
Prob. 7-44
7-41. The beam is supported by two rods AB and CD that
have cross-sectional areas of 12 mm2 and 8 mm2 , respectively.
If d = 1 m, determine the average normal stress in each rod.
7-42. The beam is supported by two rods AB a nd CD that
have cross-sectional areas of 12 mm 2 and 8 mm 2, respectively.
Determine the position d of the 6-kN load so that the average
normal stress in each rod is the same.
7-45. The plastic block is subjected to an axial compressive
force of 600 N. Assuming that the caps at the top and bottom
distribute t he load uniformly throughout the block,
determine Uhe average normal and average shear stress
acting along section a-{I.
a
150mm
a
6 kN
-d-!
1·"---1..,.....-1-f
A l .._..~~~~~~~~~----'I C
1 - - -- - - 3
J---+---+-11
Zfso
m- - ----1
Probs. 7-4V42
~O mm' 50 mn'1
I
600N
Prob. 7-45
mm
7.5
7-46. The two steel members arc joined together using a
30° scarf weld. Determine the average normal and average
shear stress resisted in the plane of the weld.
34 5
A VERAGE SHEAR STRESS
7-49. The two members used in the construction of an
aircraft fuselage are joined together using a 300 fish·mouth
weld. Determine the average normal and average shear
stress on the plane of each weld. Asswne each inclined
plane supports a horizontal force of 400 lb.
15kN
1.5 in.
30"
I L..1....L..
30"
0.
800lb
Prob. 7-49
20 mm
40 mn1-
7-50. The 2-Mg concrete pipe has a center of mass at
point G. If it is suspended from cables AB and AC,
determine the average normal stress in the cables. The
diameters of A 8 and AC are 12 mm and 10 mm. respectively.
15kN
Prob. 7-46
7-47. The bar has a cross-sectional area of 400(1o-6) m2 . If
it is subjected to a triangular axial distributed loading along
its length which is 0 at x = 0 and 9 kN/m at x = 1.5 m, and
to two concentrated loads as shown, determine the average
normal stress in the bar as a function of x for 0 < x < 0.6 m.
7-51. The 2-Mg concrete pipe has a center of mass at
point G. If it is suspended from cables AB and AC,
determine the diameter of cable AB so that the average
normal stress in this cable is the same as in the
10-mrn-diameter cable AC.
*7-48. The bar has a cross-sectional area of 400(10-6) m2 .
Hit is subjected to a uniform axial distributed loading along
its length of 9 kN/m, and to two concentrated loads as
shown, determine the average normal stress in the bar as a
function of x for 0.6 m < x < 1.5 m.
8 kN
-..
___..
--+-
___.. ____. _ . .
~
4 kN
l::::=======~~----...,,,,,,:1---+-
-:-1- -- If- -o.6 m
---
0.9 m
Probs. 7-47/48
- - -1
Probs. 7-50/51
346
CHAPTER
7
S TR ES S AND STRAIN
7. 6
ALLOWABLE STRESS DESIGN
To ensure the safety of a structural or mechanical member, it is necessary
to restrict the applied load to one that is less than the load the member
can fully support. There are many reasons for doing this.
• The intended measurements of a structure or machine may not be
exact, due to errors in fabrication or in the assembly of its component
parts.
• Unknown vibrations, impact, or accidental loadings can occur that
may not be accounted for in the design.
• Atmospheric corrosion, decay, or weathering tend to cause materials
to deteriorate during service.
• Some materials, such as wood, concrete, or fiber-reinforced
composites, can show high variability in mechanical properties.
One method of specifying the allowable load for a member is to use a
number called the factor of safety (F.S.). It is a ratio of the failure load
Frail to the allowable load Fallow,
Cranes are often support ed using
bearing pads to give them stability. Care
must be taken not to crush the
supporting surface, due to the large
bearing stress developed between the
pad and the surface.
F.S.
=
Frail
(7- 5)
Fallow
Here Frail is found from experimental testing of the material.
If the load applied to the member is linearly related to the stress
developed within the member, as in the case of a = N /A and
Tavg = V /A, then we can also express the factor of safety as a ratio of the
failure stress afail (or Trail) to the allowable stress a, 110w (or Tallow). Here the
area A will cancel, and so,
F.S.
=
Ufail
Uauow
(7-6)
F.S.
=
Tfa;1
Ta now
(7- 7)
or
7.6
Specific values of F.S. depend on the types of materials to be used
and the intended purpose of the structure or machine, while accounting
for the previously mentioned uncertainties. For example, the F.S. used
in the design of aircraft-or space-vehicle components may be close to 1
in order to reduce the weight of the vehicle. Or, in the case of a n uclear
power plant, the factor of safety for some of its components may be as
high as 3 due to uncertainties in loading or material behavior. Whatever
the case, the factor of safety or the allowable stress for a specific case
can be found in design codes and engineering handbooks. Design that
is based on an allowable stress limit is called allowable stress design
(ASD). Using this method will ensure a balance between both public
and environmental safety on the one hand and economic considerations
on the other.
Simple
Connections.
By making simplifying assumptions
regarding the behavior of the material, the equations a = N /A and
Tavg = V /A can often be used to analyze or design a simple connection
or mechanical element. For example, if a member is subjected to normal
force at a section, its required area at the section is determined from
N
A = --
347
A LLOWABLE STRESS DESIGN
p
~1
8
(ub)• .,..
Assumed uniform I
normal stress _ __,
distribution
p
A=---
~
The area of the colu mn base plate 8 is detem1ined
Crom the allowable bearing stress for the concrete.
(7-8)
a allow
or if the section is subjected to an average shear force, then the required
area at the section is
v
A = --
(7-9)
Tauow
Three examples of where the above equations apply are shown in
Fig. 7-21. The first figure shows the normal stress acting on the bottom of
a base plate. This compressive stress caused by one surface that bears
against another is often called bearing stress.
Assumed uniform shear stress
The embedded length I of this rod in concrete
can be determined using the allowable shear
stress of the bonding glue.
V=Px
Assumed uniform
p
p
A= - T allow
The area of the bolt for this lap joint
is determined from the shear stress,
which is largest between the plates.
'-......p
Fig. 7- 21
348
CHAPTER
7
STRESS AND STRAIN
IMPORTANT POINT
• Design of a member for strength is based on selecting an
allowable stress that will enable it to safely support its intended
load. Since there are many unknown factors that can influence
the actual stress in a member, then depending upon the
intended use of the member, a factor of safety is applied to
obtain the allowable load the member can support.
PROCEDURE FOR ANALYSIS
When solving problems using the average normal and shear stress
equations, a careful consideration should first be made as to choose
the section over which the critical stress is acting. Once this section
is determined, the member must then be designed to have a
sufficient area at the section to resist the stress that acts on it. This
area is determined using the following steps.
Internal Loading.
• Section the member through the area and draw a free-body
diagram of a segment of the member. The internal resultant
force at the section is then determined using the equations of
equilibrium.
Required Area.
• Provided the allowable stress is known or can be determined,
the required area needed to sustain the load at the section is
then determined from A = P/ aauow or A = V /Tallow·
Appropriate factors of safety must be
considered when designing cranes and
cables used to transfer heavy loads.
7.6
EXAMPLE
349
ALLOWABLE STRESS DESIGN
7.10
-
-
The control arm is subjected to the loading shown in Fig. 7- 22a.
Determine to the nearest in. the required diameters of the steel pins
at A and C if the factor of safety for shear is F.S. = 1.5 and the failure
shear stress is Tea;1 = 12 ksi.
l
A
B
SOLUTION
8 in.
A free-body diagram of the arm is shown in Fig. 7- 22b.
For equilibrium we have
Single
shear
Pin Forces.
~+lMe = O;
FA 8 (
8 in.) - 3 kip ( 3 in. ) - 5 kip ( ~) ( 5 in. )
FAB =
~ IF., = O;
+ jlFy
=
3 kip
f· '
·~3
in. ;1 2 in. -
+ 5 kip ( ~)
- 3 kip - Cx
Cy - 3 kip - 5 kip ( ~)
0;
c
=
=
Cx
0
=
Double
shear
1 kip
(a)
Cy = 6 kip
0
i
3 kip
The pin at C resists the resultant force at C, which is
Fe
=
Y ( 1 kip ) 2 + ( 6 kip ) 2 =
Allowable Shear Stress.
F .S.
=
We have
, 1.5
Tfail.
12 ksi;
=
Tallow
Pin A.
6.083 kip
8 ksi
Tallow =
Tallow
8 in.
This pin is subjected to single shear, Fig. 7- 22c, so that
_/dA) 2
v
A = - -·,
011
\
Tallow
2
=
3kip
8kip/ in2 ;
dA
Use
dA
=
0.691 in.
3.
=
Ans.
410.
~r
1-3 in. -
Since this pin is subjected to double shear, a shear force of
3.041 kip acts over its cross-sectional area between the arm and each
supporting leaf for the pin, Fig. 7- 22d. We have
Pin C.
A
=
__!'.:____.
Tallow'
Use
_(
de)2
11
\
2
.
=
3.041 kip_
8 kip/in2 '
c,,
2 in. -
3 kip
(b}
de = 0.696 in.
3 .
d e = 410.
Ans.
6.082 kip
~041kip
3.041 kip
Pin at C
(d}
Fig. 7-22
3 kip
~
3 kip
Pin at A
(c)
5 kip
350
CHAPTER
EXAMPLE
-
7
STRESS AND STRAIN
7 .11
-
Determine the largest load P that can be applied to the bars of the lap
joint shown in Fig. 7- 23a. The bolt has a diameter of 10 mm and an
allowable shear stress of 80 MPa. Each plate has an allowable tensile
stress of 50 MPa, an allowable bearing stress of 80 MPa, and an allowable
shear stress of 30 MPa.
SO mm
SOLUTION
To solve the problem we will determine P for each possible failure
condition; then we will choose the smallest value of P. Why?
p
(a)
Failure of the Plate in Tension.
If the plate fails in tension, it will
do so at its smallest cross section, Fig. 7- 23b.
I
6
2 p
50(lO)
N/m - 2(0.02 m)(0.015 m)
SO mm
I
p
P = 30kN
20mm 15mm
2
2
Failure of the Plate by Bearing. A free-body diagram of the top plate,
Fig. 7- 23c, shows that the bolt will exert a complicated distribution of
stress on the plate along the curved central area of contact with the bolt.*
To simplify the analysis for small connections having pins or bolts
such as this, design codes allow the projected area of the bolt to be
used when calculating the bearing stress. Therefore,
Failure of plate in tension
(b)
Actual stress
distribution
Assumed uniform
stress distribution
p
p
8 0(lO6)N/m2 -- (0.01 m)(0.015
m)
P = 12 kN
p
Failure of plate in bearing caused by bolt
(c)
Fig. 7- 23
*l11e material strength of a bolt or pin is generally greater than that of the plate material,
so bearing failure of the member is of greater concern.
7 .6
p
p
V= 2
Failure of plate by shear
(d}
Failure of the Plate by Shear.
There is the possibility for the bolt
to tear through the plate along the section shown on the free-body
diagram in Fig. 7- 23d. Here the shear is V = P /2, and so
6
2 P/2
30(lO) N/m - (0.02 m)(0.015 m)
P = 18 kN
Failure of the Bolt by Shear.
The bolt can fail in shear along the plane
between the plates. The free-body diagram in Fig. 7- 23e indicates that
V = P,so that
80(106) N/m2
=
p
7r(0.005 m) 2
P = 6.28kN
Comparing the above results, the largest allowable load for the
connections depends upon the bolt shear. Therefore,
P = 6.28kN
Ans.
p
V=P
Failure of bolt by shear
(e)
Fig. 7-23 (cont.)
ALLOWABLE STRESS DESIGN
3 51
352
CHAPTER
EXAMPLE
-
7
STRESS AND STRAIN
7 .12
-
The suspender rod is supported at its end by a fixed-connected circular
disk as shown in Fig. 7- 24a. If the rod passes through a 40-mm-diameter
hole, determine the minimum required diameter of the rod and the
minimum thickness of the disk needed to support the 20-kN load. The
allowable normal stress for the rod is aauow = 60 MPa, and the allowable
shear stress for the disk is Tallow = 35 MPa.
-1 40mm l-
d
20kN
l 20kN
(a)
(b}
Fig. 7-24
SOLUTION
By inspection, the axial force in the rod is 20 kN.
Thus the required cross-sectional area of the rod is
Diameter of Rod.
N
A =
1T
4
,
<Ta llow
d2
20(103 ) N
=-----60(106) N/m2
so that
d = 0.0206 m = 20.6 mm
Ans.
As shown on the free-body diagram in Fig. 7- 24b,
the material at the sectioned area of the disk must resist shear stress
to prevent movement of the disk through the hole. If this shear stress
is assumed to be uniformly distributed over the sectioned area, then,
since V = 20 kN, we have
Thickness of Disk.
v
A = -·,
'Tallow
20(103 ) N
27T(0.02 m) (t) - 35(106) N/m2
t = 4.55(10- 3 ) m = 4.55 mm
Ans.
7.6
EXAMPLE
A LLOWABLE STRESS D ESIGN
7 .1_:J
The shaft shown in Fig. 7-25a is supported by the collar at C, which is
attached to the shaft and located on the right side of the bearing at B.
Determine the largest value of P for the axial forces at E and F so that
the bearing stress on the collar does not exceed an allowable stress of
(ub)a11ow = 7S MPa and the average normal stress in the shaft does not
exceed an allowable stress of (u,)auow = SS MPa.
~ r-20mm
~~~~)eP~I~~
._
ja
1
80mm
A
60mm
~"==;;;;;;;;;;;;;;;;~P~~ 3P
2P ~·
C
E
F
E
C
(b)
(a)
Axial
Force
~~1:===========:--~£
F
C
Pos ition
(c)
Fig. 7-25
SOLUTION
To solve the problem we will determine P for each possible failure
condition. The n we will choose the smallest value. Why?
Normal Stress. Using the method of sections, the axial load within
region FE of the shaft is 2P, whereas the largest axial force, 3P , occurs
within region EC, Fig. 7-25b. The variation of the internal loading is
clearly shown on the normal-force diagram, Fig. 7- 25c. Since the crosssectionaJ area of the e ntire shaft is constant, region FC is subjected to
the maximum averge normal stress. Applying Eq. 7-8, we have
N
3P
2
3
A = Uanow'
7T (0.0 m ) = SS ( la6 ) N / m2
P = Sl.8 kN
Ans.
Bearing Stress. As shown o n the free-body diagram in Fig. 7-2Sd,
the collar at C must resist the load of 3P, which acts over a bearing area
of Ab= [7r(0.04 m) 2 - 7T(0.03 m)2) = 2.199(10- 3) m2. Thus,
~
N
A = Uallow'
2. 199( 10
2 _
W
) m - 75(106) N/m2
P
SS.O kN
By comparsion, the largest load that can be applied to the shaft is
P = 51.8 kN, since any load larger than this will cause the allowable
normal stress in the shaft to be exceeded.
NOTE: Here we have not considered a possible shear failure of the
collar as in E xample 7 .12.
=
3P ~~
c
(d)
353
354
CHAPTER
7
S TR ES S AND STRAIN
FUNDAMENTAL PROBLEMS
F7-13. Rods AC and BC are used to suspend the 200-kg
mass. If each rod is made of a material for which the average
normal stress can not exceed 150 MPa, determine the
minimum required diameter of each rod to the nearest mm.
F7-15. Determine the maximum average shear stress
developed in each 3/ 4-in.-diameter bolt.
10 kip
5 kip
Prob.F7-15
Prob.F7-13
F7-14. The pin at A has a diameter of 0.25 in. If it is
subjected to double shear, determine the average shear
stress in the pin.
F7-16. If each of the three nails has a diameter of 4 mm
and can withstand an average shear stress of 60 MPa,
determine the maximum allowable force P that can be
applied to the board.
. 1--2ft - - 1 - -2 ft - I
6001b
3 ft
•
Prob.F7-14
Prob.F7-16
7.6
F7-17. The strut is glued to the horizontal member at
surface AB. If the strut has a thickness of25 mm and the glue
can withstand an average shear stress of 600 kPa, determine
the maximum force P that can be applied to the strut.
355
ALLOWABLE STRESS DESIGN
F7-19. If the eyebolt is made of a material having a yield
stress of uy = 250 MPa, determine the minimum required
diameter d of its shank. Apply a factor of safety F.S. = 1.5
against yielding.
d
30kN
Prob. F7-19
Prob. F7-17
F7-18. Determine the maximum average shear stress
developed in the 30-mm-diameter pin.
F7-20. If the bar assembly is made of a material having a
yield stress of uy = 50 ksi, determine the minimum required
dimensions h 1 and hi to the nearest 1/8 in. Apply a factor
of safety F.S. = 1.5 against yielding. Each bar has a
thickness of 0.5 in.
30kN
15 kip
30 kip
40 kN
Prob. F7-18
c
B
15 kip
Prob. F7-20
A
356
7
CHAPTER
S TR ES S AND STRAIN
F7-21. Determine the maximum force P that can be
applied to the rod if it is made of material having a yield
stress of uy = 250 MPa. Consider the possibility that failure
occurs in the rod and at section a-a. Apply a factor of safety
of F.S. = 2 against yielding.
F7-23. If the bolt head and the supporting bracket are
made of the same material having a failure shear stress of
'Tra;i = 120 MPa, determine the maximum allowable force P
that can be applied to the bolt so that it does not pull
through the plate. Apply a factor of safety of F.S. = 2.5
against shear failure.
1-80mm-I
a
T
75mm
p
40mm
J
- 40mm
Section a-a
p
Prob.F7-21
Prob.F7-23
F7-22. The pin is made of a material having a failure shear
stress of 'Tra;i = 100 MPa. Determine the minimum required
diameter of the pin to the nearest mm. Apply a factor of
safety of F.S. = 2.5 against shear failure.
F7-24. Six nails are used to hold the hanger at A against
the column. Determine the minimum required diameter of
each nail to the nearest 1/16 in. if it is made of material
having Tra;1 = 16 ksi. Apply a factor of safety of F.S. = 2
against shear failure.
3001b/ft
M
80 kN
A
1 - - - - - - -9 ft - - - - - - 1
Prob.F7-22
Prob.F7-24
7.6
357
A LLOWABLE STRESS DESIGN
PROBLEMS
*7-52. If A and 8 are both made of wood and are ~in.
thick, detennine to the nearest ~in. the smallest dimension
h of the vertical segment so that it does not fail in shear. The
allowable shear stress for the segment is Ta11o.. = 300 psi.
7- 54. The connection is made using a bolt and nut and two
washers. If the allowable bearing stress of the washers on the
boards is (ub)a11ow = 2 ksi. and the allowable tensile stress
within the bolt shank S is (u,)a11ow = 18 ksi. determine the
maximum allowable tension in the bolt shank. The bolt shank
has a diameter of 0.31 in.. and the washers have an outer
diameter of 0.75 in. and inner diameter (hole) of 0.50 in.
Prob. 7-54
Prob. 7-52
7-53. The lever is attached to the shaft A using a key that
has a width d and length of 25 mm.Uthe shaft is fixed and a
vertical force of 200 N is applied perpendicular to the
handle, detennine the dimension d if the allowable shear
stress for the key is T311ow = 35 MPa.
a
7- 55. The tension member is fastened together using rwo
bolts, one on each side of the member as shown. Each bolt
has a diameter of 0.3 in. Determine the maximum load P
that can be applied to the member if the allowable shear
stress for the bolts is Tallow = 12 ksi and the allowable
average normal stress is Uauow = 20 ksi.
(I
11'
A
- 2 0 mm
SOOmm
i
200N
Prob. 7-53
p
f
A&
I.
Prob. 7-55
j
.. p
358
CHAPTER
7
S TR ES S AND STRAIN
*7-56. The steel swivel bushing in the elevator control of
an airplane is held in place using a nut and washer as shown
in Fig. (a). Failure of the washer A can cause the push rod
to separate as shown in Fig. (b). If the maximum average
shear stress is Tmax =21 ksi, determine the force F that must
be applied to the bushing. The washer is 116 in. thick.
~-21::ZOJ.~ •F-~
....
7-58. Determine the size of square bearing plates A ' and
B ' required to support the loading. Take P = 1.5 kip.
Dimension the plates to the nearest! in. The reactions at the
supports are vertical and the allowable bearing stress for the
plates is (ub)auaw = 400 psi.
7-59. Determine the maximum load P that can be
applied to the beam if the bearing plates A ' and B' have
square cross sections of 2 in. x 2 in. and 4 in. x 4 in.,
respectively, and the allowable bearing stress for the
material is ( ub)auaw = 400 psi.
,:
(b)
(a)
Prob. 7-56
3 kip
2kip
2 kip
2 kip
5 ft - i i - 5 ft - - 5 ft - ! - 7.5 ft- p
7-57. The spring mechanism is used as a shock absorber
for a load applied to the drawbar AB. Determine the force
in each spring when the 50-kN force is applied. Each spring
is originally unstretched and the drawbar slides along the
smooth guide posts CG and EF. The ends of all springs are
attached to their respective members. Also, what is the
required diameter of the shank of bolts CG and EF if the
allowable stress for the bolts is uauaw = 150 MPa?
B
Probs. 7-58/59
*7-60. Determine the required diameter of the pins at A
and B to the nearest 1~ in. if the allowable shear stress for the
material is Tauaw = 6 ksi. Pin A is subjected to double shear,
whereas pin B is subjected to single shear.
k = 80kN/m
c
A
E
H
A
B
3 kip
k' = 60kN/m
8 ft
200mm
200mm
D
50kN
Prob. 7-57
•
L6ft~-6ft-1
Prob. 7-60
D
7.6
7-61. If the allowable tensile stress for wires AB and AC is
uano,. = 200 MPa, determine the required diameter of each
wire if the applied load is P = 6 kN.
ALLOWABLE STRESS DESIGN
359
*7-64. Determine the required diameter of the pins at
A and B if the allowable shear stress for the material is
ranow = 100 MPa. Both pins are subjected to double shear.
7-62. If the allowable tensile stress for wires AB and AC is
u8110w = 180 MPa, and wire AB has a diameter of 5 mm and
AC has a diameter of 6 mm, determine the greatest force P
that can be applied to the chain.
2 kN/m
"A
3m
0
c
Prob.7-64
p
Probs. 7-6V62
7-63. The cotter is used to hold the two rods together.
Determine the smallest thickness t of the cotter and the
smallest diameter d of the rods. All parts are made of steel
for which the failure normal stress is uran = 500 MPa and
the failure shear stress is Trail = 375 MPa. Use a factor of
safety of (F.S.), = 2.50 in tension and (F.S.), = 1.75 in shear.
7-65. The steel pipe is supported on the circular base plate
and concrete pedestal. If the thickness of the pipe is
1 = 5 mm and the base plate has a radius of 150 mm,
determine the factors of safety against failure of the steel
and concrete. The applied force is 500 kN, and the normal
failure stresses for steel and concrete arc (ur3 ;1)" =350 MPa
and (ur3 ;1)con = 25 MPa, respectively.
1
SOOkN
d
.
d
'
. ..
.•
30kN
Prob. 7-63
Prob. 7-65
360
CHAPTER
7
S TR ES S AND STRAIN
7-66. The boom is supported by the winch cable that has a
diameter of 0.25 in. and an allowable normal stress of
uauow = 24 ksi. Determine the greatest weight of the crate
that can be supported without causing the cable to fail if
<f> = 30°. Neglect the size of the winch.
7-69. The two aluminum rods support the vertical force
of P = 20 kN. Determine their required diameters if the
allowable tensile stress for the aluminum is uauow = 150 MPa.
7-67. The boom is supported by the winch cable that
has an allowable normal stress of uauow = 24 ksi. If it
supports the 5000 lb crate when <f> = 20°, determine the
smallest diameter of the cable to the nearest /6 in.
B
c
0
p
Prob. 7-69
Probs. 7-66/67
*7-68. The assembly consists of three disks A , B , and C
that are used to support the load of 140 kN. Determine the
smallest diameter d 1 of the top disk, the diameter d 2 within
the support space, and the diameter d3 of the hole in the
bottom disk. The allowable bearing stress for the material
is (ub)auow = 350 MPa and allowable shear stress is
Tallow = 125 MPa.
7-70. The two aluminum rods AB and AC have diameters
of 10 mm and 8 mm, respectively. Determine the largest
vertical force P that can be supported. The allowable tensile
stress for the aluminum is uauow = 150 MPa.
140 kN
di-I
A
B
c
i I-
l- d3 -I
d1
Prob. 7-68
20mm
I
lOmm
l
c
A
Q
-I
p
Prob. 7-70
7.7
7.7
DEFORMATION
Whenever a force is applied to a body, it will tend to change the body's
shape and size. These changes are referred to as deformation, and they
may be either highly visible or practically unnoticeable. For example, a
rubber band will undergo a very large deformation when stretched,
whereas only slight deformations of structural members occur when a
building is occupied by people walking about. Deformation of a body can
also occur when the temperature of the body is changed. A t ypical
example is the thermal expansion or contraction of a roof caused by
the weather.
In a general sense, the deformation of a body will not be uniform
throughout its volume, and so the change in geometry of any line
segment within the body may vary substantially along its length. H ence,
to study deformational changes in a more uniform manner, we will
consider Line segments that are very short and located in the neighborhood
of a point. Realize, however, that these changes will also depend on the
orientation of the line segment at the point. For example, a line segment
may elongate if it is oriented in one direction, whereas it may contract if
it is oriented in another direction.
No te the befo re and afte r positions of three
d iCCferent line segments on this rubber membrane
which is subjected to tension. The vertical line is
lengthened, the horizontal line is shortened, and
the inclined line changes its le ngth and rotates.
D EFORMATION
361
362
CHAPTER
7
S TR ES S AND STRAIN
7.8 STRAIN
In order to describe the deformation of a body by changes in lengths of
line segments and changes in the angles between them, we will develop
the concept of strain. Strain is actually measured by experiment, and once
the strain is obtained, it will be shown in the next chapter how it can be
related to the stress acting within the body.
p
p
Fig. 7-26
Normal Strain. If an axial load Pis applied to the bar in Fig. 7- 26, it
will change the bar's length L 0 to a length L. We will define the average
normal strain E (epsilon) of the bar as the change in its length
o(delta) = L - L 0 divided by its original length, that is
I
E avg =
L
~o Lo I
(7- 10)
The normal strain at a point in a body of arbitary shape is defined in a
similar manner. For example, consider the very small line segment ds
located at the point, Fig. 7- 27. After deformation it becomes ds', and the
change in its length is therefore ds' - ds. As ds ~ 0, in the limit the
normal strain at the point is therefore
Undeformed body
€
=
. ds' - ds
Jim - - - Ll.S-o
ds
(7- 11)
• p
p •
Deformed body
Fig. 7- 27
In both cases
E (or Eavg) is a change in length per unit length, and it is
positive when the initial line elongates, and negative when the line
contracts.
Units. As shown, normal strain is a dimensionless quantity, since it is a
ratio of two lengths. However, it is sometimes stated in terms of a ratio of
length units. If the SI system is used, where the basic unit for length is the
meter (m), then since € is generally very small, for most engineering
applications, measurements of strain will be in micrometers per meter
(µm/m) , where 1 µm = 10- 6 m. In the Foot-Pound-Second system, strain
is often stated in units of inches per inch (in.fin.), and for experimental
work, strain is sometimes expressed as a percent. For example, a normal
strain of 480(10--6) can be reported as 480(10--6) in./in., 480 µm/m , or
0.0480%. Or one can state the strain as simply 480 µ (480 "micros").
7.8
v
v
I
I \
Undeformed body
363
STRAIN
Deformed body
\
Deformed body
(a)
(b)
~~
0
Positive shear strain y
Negative shear strain y
(c)
Fig. 7-28
Shear Strain. Deformations not only cause line segments to elongate or
contract, but they also cause them to change direction. If we select two line
segments that are originally perpendicular to one another, then the dumge
in angle that occurs between them is referred to as shear strain.This angle
is denoted by y (gamma) and is always measured in radians (rad), which are
dimensionless. For example, consider the two perpendicular line segments
at a point in the block shown in Fig. 7-2&. If an applied loading causes the
block to deform as shown in Fig. 7-28b, so that the angle between the Line
segments becomes 8, the n the shear strain at the point becomes
x
(a)
1T
2
y
7T
=- 2
(7-12)
8
r.
1T
Notice that if 8 is smaller than -rr/2, Fig. 7- 2&, then the shear strain is
positive, whereas if 8 is larger than -rr /2, then the shear strain is negative.
Cartesian S
2
~y
Undeformed
element
.omponents. We can generalize our definitions
(b)
of normal and shear strain and consider the undeformed element at a
point in a body, Fig. 7-29a. Since the element's dimensions are very small,
its deformed shape will become a parallelepiped, Fig. 7- 29b. Here the
normal strains change the sides of the element to
(~ -Y...,.)
2
(I
+ E,)6t
(; - Y,,)
which produces a change in the volume of the element. And the shear
strain changes the a ngles between the sides of the element to
7T
2-
7T
'Yxy
2-
2-
( 1 + Ey)dy
Deformed
eleme nt
7T
'Yyz
(~ - 'Yy:
2
'Yxz
which produces a change in the shape of the element.
(c)
Fig. 7-29
364
CHAPTER
7
STRESS AND STRAIN
Small Strain Analysis. Most engineering design involves applications
for which only small deformations are allowed. In this text, therefore, we
will assume that the deformations that take place within a body are almost
infinitesimal. For example, the normal strains occurring within the
material are very small compared to 1, so that e << 1. This assumption
has wide practical application in engineering, and it is often referred to as
a small strain analysis. It can also be used when a change in angle, 6.8, is
small, so that sin 6.8 "" 6.8, cos 6.8 "" 1, and tan 6.8 "" 6.8.
The rubber bearing support under this
concrete bridge girder is subjected to
both normal and shear strain. The
normal strain is caused by the weight
and bridge loads on the girder, and the
shear strain is caused by the horizontal
movement of the girder due to
temperature changes.
IMPORTANT POINTS
• Loads will cause all material bodies to deform and, as a result,
points in a body will undergo displacements or changes in
position.
• Normal strain is a measure per unit length of the elongation or
contraction of a small line segment in the body, whereas shear
strain is a measure of the change in angle that occurs between two
small line segments that are originally perpendicular to one
another.
• The state of strain at a point is characterized by six strain
components: three normal strains Ex, eY' e z and three shear strains
Yxy> Yyz > Yxz· These components all depend upon the original
orientation of the line segments and their location in the body.
• Strain is the geometrical quantity that is measured using
experimental techniques. Once obtained, the stress in the body
can then be determined from material property relations, as
discussed in the next chapter.
• Most engineering materials undergo very small deformations,
and so the normal strain e << 1. This assumption of "small strain
analysis" allows the calculations for normal strain to be simplified,
since first-order approximations can be made about its size.
7.8
EXAMPLE
7.14
D etermine the average normal strains in the two wires in Fig. 7- 30 if the
ring at A moves to A'.
i---3 m--1-B
4m
-I ~
20mm
IOmm
p
Fig. 7-30
SOLUTION
Geometry.
The original length of each wire is
LAB= LAc
= Y(3m)2 +
(4m)2
= 5m
The final lengths are
LA'B
=
LA'C
= Y(3 m + 0.01
Y(3 m - 0.01 m)2 + (4 m + 0.02 m) 2
m)2
= 5.01004 m
+ (4 m + 0.02 m) 2 = 5.02200 m
Average Normal Strain.
"AB
= LA'B -
LAB
"Ac
= LA'C LA c
= 5.01004 m
- 5m
= 2.01(10- 3) m/m
Ans.
- 5m
= 4.40(l0- 3) m/m
Ans.
5m
LAB
LA c
= 5.02200 m
Sm
STRAIN
365
366
I
CHAPTER
EXAMPLE
7
7.15
STRESS AND STRAIN
I
When force P is applied to the rigid lever arm ABC in Fig. 7- 31a, the arm
rotates counterclockwise about pin A through an angle of 0.05°.
Determine the normal strain in wire BD.
'
!p
c
D
I
SOLUTION I
300mm
B
=1
A
•
Geometry. The orientation of the lever arm after it rotates about
point A is shown in Fig. 7- 31b. From the geometry of this figure,
1 -400mm - I
a = tan- 1( 400mm)
mm = 53.1301°
300
(a)
Then
<f> = 90° - a
+ 0.05° = 90° - 53.1301° + 0.05°
=
36.92°
For triangle ABD the Pythagorean theorem gives
LAD = Y(300 mm) 2 + (400 mm) 2
=
500 mm
Using this result and applying the Jaw of cosines to triangle AB' D ,
LB'D
j -400mm
=
YL~v + L~B' - 2(LAv) (LAB') cos <P
Y(500 mm) 2 + (400 mm) 2 - 2(500 mm) (400 mm) cos 36.92°
=
300.3491 mm
=
D
a
P
c
300mm
~~ o= o.os·~"':=:==~A~l
B' _
Normal Strain.
LB'D - LBv
LBv
300.3491 mm - 300 mm
mm
300
400mm(b)
=
0.00116 mm/mm
Ans.
Fig. 7-31
SOLUTION II
Since the strain is small, this same result can be obtained by approximating
the elongation of wire BD as 6.LBv, shown in Fig. 7- 31b. Here,
l!..LBD
=
OLAB
=
0.05°)
J
[( 1800 (1T rad) (400 mm)
=
0.3491 mm
Therefore,
0.3491 mm
mm
300
=
0.00116 mm/mm
Ans.
7.8
-
EXAMPLE
367
STRAIN
7 .16
-
The plate shown in Fig. 7- 32a is fixed connected along AB and !held in
the horizontal guides at its top and bottom, AD and BC. If its right side
CD is given a uniform horizontal displacement of 2 mm, determine
(a) the average normal strain along the diagonal AC, and (b) the shear
strain at E relative to the x , y axes.
SOLUTION
x
y
-- A
l
I'
I
I
I
150 mm
Part (a). When the plate is deformed, the diagonal AC becomes
AC', Fig. 7- 32b. The lengths of diagonals AC and AC' can be found
from the Pythagorean theorem. We have
AC = V (0.150 m ) 2
+ ( 0.150 m ) 2
=
1
:
B------~
c;.;...-
- - 150 mm
0.21213 m
!
E
--11- 2 mm
(a)
AC' = V(0.150m) 2 + (0.152m) 2 = 0.21355m
Therefore the average normal strain along AC is
AC' - AC
0.21355 m - 0.21213 m
-------(EAc)avg AC
0.21213 m
=
0.00669 mm/mm
Ans.
Part (b). To find the shear strain at E relative to the x and y axes,
which are 90° apart, it is necessary to find the change in the angle at£.
The angle 8 after deformation, Fig. 7- 32b, is
(8)
tan 2
8 = 90.759° = (
Fig. 7- 32
=
76mm
75 mm
; ) (90.759°)
1 00
=
1.58404 rad
Applying Eq. 7- 12, the shear strain at E is therefore the change in the
angleAED,
'Yxy =
1T
2-
1.58404 rad
=
-0.0132 rad
Ans.
The negative sign indicates that the once 90° angle becomes larger.
NOTE: If the x and y axes were horizontal and vertical at point E, then
the 90° angle between these axes would not change due to the
deformation, and so 'Yxy = 0 at point E.
368
CHAPTER
7
S TR ES S AND STRAIN
PRELIMINARY PROBLEMS
P7-7. A loading causes the member to deform into the
dashed shape. Explain how to determine the normal strains
Ecv and IEAB· The displacement
a and the lettered
dimensions are known.
P7-10. A loading causes the block to deform into the
dashed shape. Explain how to determine the strains £An,
£Ac, £ 8 c , ("YA},,,. The angles and distances between all
lettered points are known.
v\ '-r
L
B\ L
l
f L/2
'"'
A
;!i~-~-=-_=:=::::=cl!J.nr-+l8
l- L J_:~-1
-- - --
y
81------~ C
Prob. P7- 7
I
-------- ,C'
1B'
I
I
I
P7-8. A loading causes the member to deform into the
dashed shape. Explain how to determine the normal strains
Ecv and IEAB· The displacement
a and the lettered
dimensions are known.
I
I
I
I
I
I
I
I
I
..._..__
I~ _ _ ___,_I_ _ _ _ x
A
D
Prob. P7-10
~--
--
- - -2 L - - - - 1 - - L -
Prob. P7-8
P7- l l A loading causes the block to deform into the
dashed shape. Explain how to determine the strains (1'A)xy•
("y 8 )xy- The angles and distances between all lettered points
are known.
P7-9. A loading causes the wires to elongate into the
dashed shape. Explain how to determine the normal strain
IEAB in wire AB. The displacement a and the distances
between all lettered points are known.
y
c
B
,- -- -- - /09
x
2
I
'
'
I
.. ~I
\ \
- - -
y
~
Id
•:.-
A'
Prob. P7-9
Prob. P7- 11
7.8
STRAIN
369
FUNDAMENTAL PROBLEMS
F7-25. When force P is applied to the rigid arm ABC,
point B displaces vertically downward through a distance of
0.2 mm. Determine the normal strain in wire CD.
v\
- -400mm l
'1
F7-28. The triangular plate is deformed into the shape
shown by the dashed line. Determine the normal strain
developed along edge BC and the average shear strain at
corner A with respect to the x and y axes.
200mm 3imm
y
Afo,,
.
B
'
c
p
Smm
1-A- - -400 mm- - -·11·--1
--.---+~~
-~
-~-=-----------,.;,--+-1,.---~ x
---
Prob. F7-25
If the force P causes the rigid arm ABC to rotate
clockwise about pin A through an angle of 0.02°, determine
the normal strain in wires BD and CE.
F7-26.
-~-- ~ mm
300mm
l
c
Prob. F7-28
F7-29. The square plate is deformed into the shape shown
Prob. F7-26
by the dashed line. Determine the average normal strain
along diagonal AC and the shear strain at point E with
respect to the x and y axes.
F7-'1:7. The rectangular plate is deformed into the shape of a
parallelogram shown by the dashed line. Determine the average
shear strain at comer A with respect to the x and y axes.
x
y
c
D
IT
J 4mm
300mm1
~~~-----~-·
A
- - - - - - - 11 l4mm
- 300mm
Prob. F7-27
l-~ 300mm~H
3mm
3mm
Prob. F7-29
370
CHAPTER
7
S TR ES S AND STRAIN
PROBLEMS
7- 71. An air-filled rubber ball has a diameter of 6 in. If the
air pressure within the ball is increased until the diameter
becomes 7 in., determine the average normal strain in the
rubber.
7-75. The rectangular plate is subjected to the deformation
shown by the dashed lines. Determine the average shear
strain y_.,. in the plate.
*7-72. A thin strip of rubber has an unstretched length of
15 in. If it is stretched around a pipe having an outer diameter
of 5 in., determine the average normal strain in the strip.
y
7- 73. If the load P on the beam causes the end C to be
displaced 10 mm downward, determine the normal strain in
wires CE and BD.
D
-
150mm -
3mm
-,
B _l_
-
r--=-=
- -""'-==--=---------_-t
200mm
A~
-------- -,L
•
.
3mm
p
•
A
Pro b. 7-75
i
B
- - - 3m- - - l - 2m - l -2m - I
Prob. 7- 73
7- 74. The force applied at the handle of the rigid lever
causes the lever to rotate clockwise about the pin B through
an angle of 2°. Determine the average normal strain in each
wire. The wires are unstretched when the lever is in the
horizontal position.
*7- 76. The square deforms into the position shown by the
dashed lines. Determine the shear strain at each of its
corners, A , B , C, and D, relative to the x ,y axes. Side D' B '
remains horizontal.
y
__o; ____ ________~l In 3 mm
1; - - - - - - - - - - n - - -
G
ZOO mm -
AI
_L
-I
F
ZOOmm 300mm
300 mm
E
B
0
T
200mm
1
1
I
I
I
c
D
D
1
I
I
53 mm :
~-:.___9_1_.s_·_._
'
___c......
\L '
Al- - - 50 mm ---1IH I c
H
8mm
Prob. 7-74
Pro b. 7-76
7.8
7-77. The pin-connected rigid rods AB and BC are
inclined at 8 = 30" when they are unloaded. When the force
P is applied 8 becomes 30.2°. Determine the average normal
strain in wire AC.
371
STRAIN
*7--80. Determine the shear strain "Ix, at corners A and B
if the plastic distorts as shown by the dashed lines.
7--81. Determine the shear strain "Ix,• at corners D and C
if the plastic distorts as shown by the dashed lines.
p
y
12mm
i---=1_
-~ l==!--------------r-----/
4mm
3 mm
,
In
8 'mm
I
,
c :
:
--
I
I
I
I
I
I
I
!
300mlm ,/
- - --'-,-;2
t..~"""'~-=-~
-~
-~
- -~-=-=-=
-- ~.-~
' ...J..O
L ~Ou
lDwl'-A
.r
D
Prob. 7-77
1- --400 mm---I 1
5mm
Probs. 7-80/81
=
7-78. The wire AB is unstretched when 8 45°. lf a load is
applied to the bar AC. which causes 8 to become 47°,
determine the normal strain in the wire.
7-79. If a load applied to the bar AC causes point A
to be displaced to the right by an amount 6 L, determine the
norma l strain in the wire AB. Originally, 8 = 45°.
7--82. The material distorts into the d ashed position
shown. Determine the average normal strains £.r £" and the
shear strain 1'xy at A, and the average normal strain along
line BE.
7-83. 111e materia l distorts into the dashed position
shown. De te rm ine the average normal strains a long the
diagona ls AD and CF.
30mm
DH
...,-- -r . - - - ---,- ,
I
I
I
I
I
I
L
I
E I
I
/ 50mm
Probs. 7-78r79
Probs. 7-82183
372
CHAPTER
7
STRESS AND STRAIN
*7-S4. Determine the shear strain 'Yxy at corners A and B if
the plastic distorts as shown by the dashed lines.
7-85. Determine the shear strain 'Yxy at corners D and C if
the plastic distorts as shown by the dashed lines.
7-86. Determine the average normal strain that occurs
along the diagonals AC and DB.
*7-88. The triangular plate is fixed at its base. and its apex A
is given a horizontal displacement of S mm. Determine the
shear strain. -y...,.,at A.
7-89. The triangular plate is fixed at its base, and its apex A
is given a horizontal displacement of S mm. Determine the
average normal strain E, along the x axis.
7-90. 1l1e triangu lar plate is fixed at its base, and its apex A
is given a horizontal displacement of S mm. Determine the
average normal strain E,• along the x' axis.
~
y
2mm
I __!
---------------:
C '
'
'''
300m
lm
!
'
'
'
'''
!
-----------
--~=y2 mm
x
~400mm ~~
3mm
Probs. 7-84/85/86
7-87. The corners of the square plate are given the
displacements indicated. Determine the average normal
strains 1:, and "r along the x and y axes.
Probs. 7-88/89/90
7-91. The polysulfone block is glued at its top and bottom
to the rigid plates. If a tangential force, applied to the top
plate, causes the material to deform so that its sides are
described by the equation y = 3.56x114 , determine the shear
strain at the corners A and B.
y
0.2 in.
[
IOin.
Bl
-') -
0.3 in.
0.3 in. 10 in.
''
10 in.
'
c'
,,
,
lOin.--
l
I
x
P-~J=:=:=:=:=:::::J
1
y = 3.56x 114
2 in.
Al--------4~
0.2 in.
Prob. 7-87
Prob. 7- 91
x
CONCEPTUAL PROBLEMS
373
CONCEPTUAL PROBLEMS
C7- 1. Hurricane winds have caused the failure of this
highway sign. Assuming the wind creates a uniform pressure
on the sign of 2 kPa, use reasonable dimensions for the sign
and determine the resultant shear and moment at each of
the two connections where the failure occurred.
C7- 3. Here is an example of the single shear failure of a
bolt. Using appropriate free-body diagrams, explain why
the bolt failed along the section between the plates, and not
along some intermediate section such as a-a.
Prob.C7-3
Prob. C7-1
C7-4. The vertical load on the hook is 1000 lb. Draw the
appropriate free-body diagrams and determine the maximum
average shear force on the pins at A , B, and C. Note that due
to symmetry four wheels are used to support the loading on
the railing.
C7- 2. High-heel shoes can often do damage to soft wood
or linoleum floors. Using a reasonable weight and
dimensions for the heel of a regular shoe and a high-heel
shoe, determine the bearing stress under each heel if the
weight is transferred down only to the heel of one shoe.
•
11
Prob. C7-2
Prob.C7-4
374
CHAPTER
7
S TR ES S AND STRAIN
CHAPTER REVIEW
The internal loadings in a body consist
of a normal force, shear force, bending
moment, and torsional moment. They
represent the resultants of both a
normal and shear stress distribution
that act over the cross section. To obtain
these resultants, use the method of
sections and the equations of
equilibrium.
'l.F,
=0
I.F,.
0
0
=
'l.F =
'
'l.M, =
I.M,. =
0
0
Torsional
moment
T
~N Normal
force
Bending M ~--"1"-­
moment
'l.M, = 0
v
Shear
~force
F2
If a bar is made from homogeneous
isotropic material and it is subjected to
a series of external axial loads that pass
through the centroid of the cross
section, then a uniform normal stress
distribution will act over the cross
section. This average normal stress can
be determined from u = P /A , where P
is the internal axial load at the section.
The average shear stress can be
determined using Tavg = V /A , where V
is the shear force acting on the cross
section. This formula is often used to
find the average shear stress in fasteners
or in parts used for connections.
p -
-
p
u= A
Tavg
v
=A
-'avg = -v
A
p
CHAPTER REVIEW
The design of any simple connection
requires that the average stress along
any cross section not exceed an allowable
stress of uauow or 'Tallow· These values are
reported in codes and are considered
safe on the basis of experiments or
through experience. Sometimes a factor
of safety is reported provided the failure
stress is known.
Tfail
F.S. =
'Tallow
Deformation is defined as the change in
the shape and size of a body. It causes
line segments to change length and
orientation.
Normal strain Ethe change in length per
unit length of a line segment. If E is
positive, the line segment elongates. If it
is negative, the line segment contracts.
Shear strain 1' is a measure of the change
in angle made between two line
segments
that
are
originally
perpendicular to one another.
Strain is dimensionless; however, E is
sometimes reported in in.fin., mm/mm,
and 1' is in radians.
Eavg
=
ds' - ds
ds
7T
y= - - 8
2
375
376
CHAPTER
7
S TR ES S AND STRAIN
REVIEW PROBLEMS
R7-1. The beam AB is pin supported at A and supported
by a cable BC. A separa1e cable CG is used to hold up the
frame. If AB weighs 120 lb/ft and the column FC has a
weight of 180 lb /ft, determine the resultant internal loadings
acting on cross sections located at points D and £. Neglect
the thickness of both the beam and column in the
calculation.
R7- 3. Determine the required thickness of member BC and
the diameter of the pins at A and B if the allowable normal
stress for member BC is uauow = 29 ksi, and the allowable
shear stress for the pins is rauow = 10 ksi.
8 ft
t
E
4 ft
_L
G
Prob. R7- 3
Prob. R7- 1
R7-2. The long bolt passes through the 30-mm-thick plate.
If the force in the bolt shank is 8 kN, determine the average
normal stress in the shank, the average shear stress along
the cylindrical area of the plate defined by the section lines
a-a, and the average shear stress in the bolt head along the
cylindrical area defined by the section lines b-b.
*R7-4. The circular punch B exerts a force of 2 kN on the
top of the plate A. Determine the average shear stress in the
plate due to this loading.
2kN
8mm
1-Y
a
7mm
18~mb
""""':;ii:;:;:;:;;:~'~:J-~8~kN
I
b-
a
Prob. R7- 2
l2mm
30mm
Prob. R7-4
377
REVIEW PROBLEMS
R7-5. Determine the average punching shear stress the
circular shaft creates in the metal plate through section AC
and BD. Also. what is the bearing stress developed on the
surface of the plate under the shaft?
R7- 7. The square plate is deformed into the shape shown
by the dashed lines. Lr DC has a normal strain Ex = 0.004,
DA has a normal strain E1 =0.005 and at D. "Yxy = 0.02 rad,
determine the average normal strain along diagonal CA.
*R7-8. The square plate is deformed into the shape shown
by the dashed lines. lf DC has a normal strain Ex = 0.004,
DA has a normal strain e1 =0.005 and at D, "Yxy = 0.02 rad,
determine the shear strain at point £with respect to the x'
and y' axes.
40kN
y
x'
y'
c
'--600mmA' ______ _
D
- - -B'
8
-60mm-
I
I
I
I
I
600mm
1- - - - - 120 mm - -- -1
I
I
£
I
I
I
I
I
Prob. R7-5
I
...l..-~-----~LL---- x
CC'
D
Probs. R7-7/8
R7-6. The bearing pad consists of a 150 mm by 150 mm
block of aluminum that supports a compressive load of
6 kN. Determine the average normal and shear stress acting
on the plane through section a-a. Show the results on a
differential volume element located on the plane.
6kN
R7- 9. The rubber block is fixed along edge AB, and
edge CD is moved so that the vertical displacement of any
point in the block is given by v(x) = (v0 / b3) x3 . Determine
the shear strain "Yxy at points (b/ 2. a / 2) and (b, a).
y
a
v (xf) ,. 1
- -----
a
- --150 mm + --1
Prob. R7-6
---
DI
......... ...·
Prob. R7-9
i-,1
Vo
__L
CHAPTER
B
I
I
(©Tom Wang/Alamy)
Horizontal ground displacements caused by an earthquake produced fracture of
this concrete column. The material properties of the steel and concrete must
be determined so that engineers can properly design the column to resist the
loadings that caused this failure.
MECHANICAL
PROPERTIES OF
MATERIALS
CHAPTER OBJECTIVES
•
To show how stress can b e related to strain by using experimental
methods to determine the stress-strain diagram for a particular
material.
•
To discuss the properties of the stress- strain diagram for materials
commonly used in engineering.
8.1
THE TENSION AND COMPRESSION
TEST
The strength of a material depends on its ability to sustain a load without
undue deformation or failure. This strength is inherent in the material
itself and must be determined by experiment. One of the most important
tests to perform in this regard is the tension or compression test. Once
this test is performed, we can then determine the relationship between
the average normal stress and average normal strain in many engineering
materials such as metals, ceramics, polymers, and composites.
379
380
CHAPTER
d0 = 0.5
-
l
8
ME CHANICAL PROPERTIES OF MATERIALS
in.
[Lo ~ 2in.J
Fig. 8-1
Typical steel specimen with a11ached
strain gage
movable
upper
crosshead
To perform a tension or compression test, a specimen of the material is
made into a "standard" shape and size, Fig. 8-1. As shown it has a constant
circular cross section with enlarged ends, so that when tested, failure will
occur somewhere within the central region of the specimen. Before
testing, two small punch marks are sometimes placed along the specimen's
uniform length. Measurements are taken of both the specimen's initial
cross-sectional area, ~. and the gage-length distance L 0 between the
punch marks. For example, when a metal specimen is used in a tension
test, it generally has an initial diameter of d0 = 0.5 in. (13 mm) and a
gage length of Lo = 2 in. (51 mm), Fig. 8- 1. A testing machine like the
one shown in Fig. 8- 2 is then used to stretch the specimen at a very slow,
constant rate until it fails. The machine is designed to read the load
required to maintain this uniform stretching.
At frequent intervals, data is recorded of the applied load P. Also, the
elongation 8 = L - L 0 between the punch marks on the specimen may
be measured, using either a caliper or a mechanical or optical device
called an extensometer. Rather than taking this measurement and then
calculating the strain, it is also possible to read the normal strain directly
on the specimen by using an electrical-resistance strain gage, which
looks like the one shown in Fig. 8-3. As shown in the adjacent photo, the
gage is cemented to the specimen along its length, so that it becomes an
integral part of the specimen. When the specimen is strained in the
direction of the gage, both the wire and specimen will experience the
same deformation or strain. By measuring the change in the electrical
resistance of the wire, the gage may then be calibrated to directly read
the normal strain in the specimen.
---
• £ZllC[TI ...
0.
•
~
.....
load
dial
--=-
f
motor
and load
controls
Electrical-resistance
strain gage
Fig. 8-2
Fig. 8-3
8.2
8.2
THE STRESS-STRAIN DIAGRAM
381
THE STRESS-STRAIN DIAGRAM
Once the stress and strain data from the test are known, then the results
can be plotted to produce a curve called the stress-strain diagram. This
diagram is very useful since it applies to a specimen of the material made
of any size. There are two ways in which the stress- strain diagram is
normally described.
Conventional Stress-Strain Diagram. The nominal or
engineering stress is determined by dividing the applied load P by the
specimen's original cross-sectional area~· This calculation assumes that
the stress is constant over the cross section and throughout the gage
length. We have
(8- 1)
Likewise, the nominal or engineering strain is found directly from the
strain gage reading, or by dividing the change in the specimen's gage
length, 8, by the specimen's original gage length L 0 . Thus,
(8-2)
When these values of a and e are plotted, where the vertical axis is the
stress and the horizontal axis is the strain, the resulting curve is called a
conventional stress- strain diagram. A typical example of this curve is
shown in Fig. 8-4. Realize, however, that two stress- strain diagrams for a
particular material will be quite similar, but will never be exactly the
same. This is because the results actually depend upon such variables as
the material's composition, microscopic imperfections, the way the
specimen is manufactured, the rate of loading, and the temperature
during the time of the test.
From the curve in Fig. 8-4, we can identify four different regions in
which the material behaves in a unique way, depending on the amount of
strain induced in the material.
"
"· 1--------7"7--?~
racture
stress
Elastic Behavior. The initial region of the curve, indicated in light ~;1=~1--1'
orange, is referred to as the elastic region. Here the curve is a straight line
up to the point where the stress reaches the proportional limit, apl· When
the stress slightly exceeds this value, the curve bends until the stress
reaches an elastic limit. For most materials, these points are very close,
and therefore it becomes rather difficult to distinguish their exact values.
What makes the elastic region unique, however, is that after reaching ay,
if the load is removed, the specimen will recover its original shape. In
other words, no damage will be done to the material.
~::::=::::::::::::::=::::±::====::::=====±==:::;::::==t- '
strain
nee ·ing
hardening
plastic behavior
Conventional and true stress-strain diag.rant
for ductile mate rial (steel) (not to scale)
Fig. 8-4
382
CHAPTER
8
ME CHANICAL PROPERTIES OF MATERIALS
Because the curve is a straight line up to <Tp/, any increase in stress will
cause a proportional increase in strain. This fact was discovered in 1676
by Robert Hooke, using springs, and is known as Hooke's law. It is
expressed mathematically as
(8- 3)
u = Ee
Here E represents the constant of proportionality, which is called the
modulus of elasticity or Young's modulus, named after Thomas Young,
who published an account of it in 1807.
As noted in Fig. 8-4, the modulus of elasticity represents the slope of
the straight line portion of the curve. Since strain is dimensionless, from
Eq. 8- 3, E will have the same units as stress, such as psi, ksi, or pascals.
true fracture stress
u/ ~~~~~~~~~~~~~~~~---'»
fracture
stress
elastic yielding
region
elastic
ehavior
strain
hardening
plastic behavior
necking
Conventional and true stress- strain diagram
for ductile material (steel) (not to scale)
Fig. 8-4 (Repeated)
Yielding. A slight increase in stress above the elastic limit will result in
a breakdown of the material and cause it to deform permanently. This
behavior is called yielding, and it is indicated by the rectangular dark
orange region in Fig. 8-4. The stress that causes yielding is called the
yield stress or yield point, <Ty, and the deformation that occurs is called
plastic deformation. Although not shown in Fig. 8-4, for low-carbon
steels or those that are hot rolled, the yield point is often distinguished
by two values. The upper yield point occurs first, followed by a sudden
decrease in load-carrying capacity to a lower yield point. Once the yield
point is reached, then as shown in Fig. 8-4, the specimen will continue to
elongate (strain) without any increase in Load. When the material behaves
in this manner, it is often referred to as being perfectly plastic.
8.2
THE STRESS-STRAIN DIAGRAM
383
Strain Hardening. When yielding has ended, any load causing an
increase in stress will be supported by the specimen, resulting in a curve
that rises continuously but becomes flatter until it reaches a maximum
stress referred to as the ultimate stress, a,.. The rise in the curve in this
manner is called strain hardening, and it is identified in Fig. 8-4 as the
region in Light green.
Necking. Up to the ultimate stress, as the specimen elongates, its
cross-sectional area will decrease in a fairly 11nifom1 manner over the
specimen's enti re gage length. However, just after reaching the ultimate
stress, the cross-sectional area will then begin to decrease in a localized
region of the specimen, and so it is here where the stress begins to
increase. As a result, a constriction or ··neck" tends to form with further
elongation, Fig. 8-Sa. This region of the curve due to necking is indicated
in dark green in Fig. 8-4. Here the stress- strain diagram tends to curve
downward until the specimen breaks at the fracture stress, a1, Fig. 8-Sb.
True Stress· Strain Diagram. Instead of always using the original
cross-sectional area A 0 and specimen length L 0 to calculate the
(engineering) stress and strain, we could have used the actual crosssectional area A and specimen length Lat the instant the load is measured.
The values of stress and strain found from these measurements are called
true stress and true strain, and a plot of their values is called the true
stress-strain diagram. When this diagram is plotted, it has a form shown
by the upper blue curve in Fig. 8-4. Note that the conventional and true
a-E diagrams are practically coincident when the strain is small.
The differences begin to appear in the strain-hardening range, where the
magnitude of strain becomes more significant. From the conventional
a-E diagram, the specimen appears to support a decreasing stress (or
load), since Ao is constant, a = N / Ao. In fact, the true a -E diagram shows
the area A within the necking region is always decreasing until fracture,
a[, and so the material actually sustains increasing stress, since a = N / A.
Although there is this divergence between these two diagrams, we can
neglect this effect since most engineering design is done only within the
elastic range. This will generally restrict the deformation of the material
to very small values, and when the load is removed the material will
restore itself to its original shape. The conventional stress- strain diagram
can be used in the elastic region because the true strain up to the elastic
Limit is small enough, so that the error in using the engineering values of
a and Eis very small (about 0.1°/o) compared with their true values.
•
Necking
• •
(a)
goe
Failure of a
ductile material
(b)
Fig. 8-5
$1•
Typica l necking pattern
which has occurred on 1his
steel specimen just before
fracture.
This steel specimen clearly shows the necking
that occurred just before the specimen failed.
This resulted in the formation of a
"curxone·· shape al the fracture location,
which is characteristic of ductile materials.
384
CHAPTER
8
ME CHANICAL PROPERTIES OF MATERIALS
Steel. A typical conventional stress- strain diagram for a mild steel
specimen is shown in Fig. 8--6. In order to enhance the details, the elastic
region of the curve has been shown in green using an exaggerated strain
scale, also shown in green. Following this curve, as the load (stress) is
increased, the proportional limit is reached at <Tp/ = 35 ksi (241 MPa),
where Ep/ = 0.0012 in.Jin. When the load is further increased, the stress
reaches an upper yield point of (<Ty) 11 = 38 ksi (262 MPa), followed by a
drop in stress to a lower yield point of (<Ty) 1 = 36 ksi (248 MPa). The
end of yielding occurs at a strain of E y = 0.030 in.Jin., which is 25 times
greater than the strain at the proportional limit! Continuing, the specimen
undergoes strain hardening until it reaches the ultimate stress of
u 11 = 63 ksi (434 MPa); then it begins to neck down until fracture
occurs, at <Ft = 47 ksi (324 MPa). By comparison, the strain at failure,
Et = 0.380 in.Jin., is 317 times greater than Ep 1!
Since <Tp/ = 35 ksi and EpJ = 0.0012 in./ in., we can determine the
modulus of elasticity. From Hooke's Jaw, it is
E
= <Tp/ =
Ep/
35 ksi
0.0012 in.Jin.
=
29 ( 103 ) ksi
Although steel alloys have different carbon contents, most grades of
steel, from the softest rolled steel to the hardest tool steel, have about
this same modulus of elasticity, as shown in Fig. 8- 7.
u (ksi)
u (ksi)
180
spring steel
(1 % carbon)
160
140
120
100
80
20
60
10
(0.050 0.10 \
0.001
Ey = 0.030
0.20
0.002
~p1= 0.0012
0.30
0.003
J
40
E
0.40
0.004
(in.f in.)
20
hard steel
(0.6% carbon)
heat treated
machine steel
(0.6% carbon)
structural steel
(0.2% carbon)
, _,_- soft steel
(0.1 % carbon)
Et= 0.380
' - -- - ' - - - ' - - - ' - - - - ' - - ' - - E
Stress- strain diagram for mild steel
Fig. 8-6
0.002 0.004 0.006 0.008 0.01
Fig. 8-7
(in.fin.)
8.3
8. 3
STRESS-STRAIN BEHAVIOR OF
Ducm.E AND BRITTLE M ATERIALS
385
STRESS-STRAIN BEHAVIOR OF
DUCTILE AND BRITTLE MATERIALS
Materials can be classified as either being ductile or brittle, depending on
their stress-strain characteristics.
Ductile Ma .e
Any material that can be subjected to Large
strains before it fractu res is called a ductile material. Mild steel, as
discusse d previously, is a typical example. Engineers o ften choose ductile
materials for design because these materials are capable of absorbing
shock or energy, and if they become overloaded, they will usually exhibit
large deformation before failing.
O ne way to specify the ductility of a material is to report its percent
elongation or pe rce nt red uction in area at the time of fracture. The
percent elongation is the specimen's fracture strain expressed as a
percent. Thus, if the specimen's o riginal gage length is L 0 and its len gth at
fracture is L 1, then
Pe rcent e lo ngation
=
L1 - L o
---'--- (100%)
Lo
(~)
For example, as in Fig. 8-6, since e1 = 0.380, this value would be 38% for
a mild steel specimen.
The percent reduction in area is another way to specify ductility. It is
defined within the region of necking as follows:
Pe rcent reduction of area =
Ao - At
Ao (lOOo/o)
(8-5)
H ere Ao is the specime n's original cross-sectional area and A1 is the area
of the neck at fracture. Mild steel has a typical value of 60%.
Besides steel, other me tals such as brass, molybdenum, and zinc may
also exhibit ductile stress~train characteristics similar to steel, wh ereby
they undergo e lastic stress-strain behavior, yielding at constant stress,
strain hardening, and finally necking until fracture. In most metals and
some plastics, however, constant yielding will not occur beyond the elastic
range. One metal where this is the case is aluminum, Fig. 8-8. Actually,
this metal often does not have a well-defined yield point, and consequently
it is standard practice to define a yield strength using a graphical procedure
called the offset method. Normally for structural design a 0.2% strain
(0.002 in.Jin.) is chosen, a nd from this point on the e axis a line paralle l to
the initial straight line portion o f the stress- strain diagram is drawn. The
point where this line inte rsects the curve defin es the yield strength. From
the graph, the yie ld strength is uys = 51 ksi (352 MPa).
tT
(ksi)
60
uys = 51
50 ~----=--,......
40
30
20
10
"---'--"-'---'--'---'--'-'--JL........I._
lo.0o2I
0.005
e (in.fin.)
0.010
(0.2% offset)
Yie ld stre ngth for an aluminum a lloy
Fig. 8-8
386
CHAPTER
8
ME CHANICAL PROPERTIES OF MATERIALS
u (ksi)
2.0
1.5
u (ksi)
1.0
B
0.5
- 0.06 - 0.05 - 0.04 - 0.03 - 0.02
--~-~--~-~----~--!>--~- E
0.01
u - E diagram
(in.f in.)
for natural rubber
Fig. 8-9
- 60
- 80
- 100
- 120
c
u - E diagram
for gray cast iron
Fig. 8-10
Concrete used for sLructural purposes
must be tested in compression to be
sure it reaches its ultimate design
stress after curing for 30 days.
Realize that the yield strength is not a physical property of the material,
since it is a stress that causes a specified permanent strain in the material.
In this text, however, we will assume that the yield strength, yield point,
elastic limit, and proportional limit all coincide unless otherwise stated.
An exception would be natural rubber, which in fact does not even have
a proportional limit, since stress and strain are not linearly related.
Instead, as shown in Fig. 8-9, this material, which is known as a polymer,
exhibits nonlinear elastic behavior.
Wood is a material that is often moderately ductile, and as a result it is
usually designed to respond only to elastic loadings. The strength
characteristics of wood vary greatly from one species to another, and for
each species they depend on the moisture content, age, and the size and
arrangement of knots in the wood. Since wood is a fibrous material, its
tensile or compressive characteristics parallel to its grain will differ
greatly from these characteristics perpendicular to its grain. Specifically,
wood splits easily when it is loaded in tension perpendicular to its grain,
and consequently tensile loads are almost always intended to be applied
parallel to the grain of wood members.
8.3
STRESS-STRAIN BEHAVIOR OF
Ducm.E AND BRITTLE M ATERIALS
387
u (ksi)
(u,)mu -- 0.4 2
- 0.0025
-0.0015
-0.(XX)S "--
E
(in.fm.)
0 0.0005
-2
-4
- -- = ------Ir-- (u,)max = 5
Tension failure of
a brinle material
(a)
Compression causes
material to bulge out
(b)
Fig. 8-11
-6
u-E diagram for typical concrete mix
Fig. 8-12
Brittle Materials. Materials that exhibit little or no yielding before
failure are referred to as brittle materials. Gray cast iron is an example,
having a stress~strai n diagram in tension as shown by the curve AB in
Fig. 8-10. Here fracture at <r1 = 22 ksi (152 MPa) occurred due to a
microscopic crack, which then spread rapidly across the specimen, causing
complete fracture. Since the appearance of initial cracks in a specimen is
quite random, brittle mate rials do not have a well-defined tensile fracture
stress. Instead the average fracture stress from a set of observed tests is
generally reported. A typical failed specimen is shown in Fig. 8-lla.
Steel rapidly loses its strength when
Compared with their be havior in tension, brittle materials exhibit a
heated. For this reason engineers often
much higher resista nce to axial compression, as evidenced by segment
require main structural members to be
AC of the gray cast iron curve in Fig. 8-10. For this case any cracks or
insulated in case of fire.
imperfections in the specimen tend to close up, and as the load increases
the material will gene rally bulge or become barrel shaped as the strains u (ksi)
become larger, Fig. 8-11 b.
40" F
9
Like gray cast iron, concrete is classified as a brittle material, and it
also has a low stre ngth capacity in tension. The characteristics of its 8
stress- strain diagram de pend primarily on the mix of concrete (water, 7
sand, gravel, and cement) and the time and temperature of curing.
A typical example of a ·'complete" stress- strain diagram for concrete is 6
given in Fig. 8-12. By inspection, its maximum compressive strength is 5
about 12.5 times greater than its tensile Strength, (Uc) max = 5 ksi 4
160" F
(34.5 MPa) versus ( <r,) max = 0.40 ksi (2.76 MPa). For this treason,
3
concrete is almost always reinforced with steel bars or rods whenever it
2
is designed to support tensile loads.
It can generally be stated that most materials exhibit both ductile and l
brittle behavior. For example, steel has brittle behavior when it contains
" --'--..1---1.--'---'' - - - - ' - - E (in.fin.)
a high carbon content, and it is ductile when the carbon content is
0.0 I 0.02 0.03 0.04 0.05 0.06
reduced. Also, at low te mperatures materials become harder and more
u-E diagrams for a melhacrylate plastic
brittle, whereas whe n the temperature rises they become softer and more
Fig. 8-13
ductile. This effect is shown in Fig. 8-13 for a methacrylate plastic.
388
CHAPTER
8
ME CHAN I CAL PROPERTIES OF MATERIALS
Stiffness. The modulus of elasticity is a mechanical property that
indicates the stiffness of a material. Materials that are very stiff, such as
steel, have large values of £(£5 1 = 29(103) ksi or200 GPa], whereas spongy
materials such as vulcanized rubber have low values (Er = 0.10 ksi or
0.69 MPa]. Values of E for commonly used engineering materials are
often tabulated in engineering codes and reference books. Representative
values are also listed on the inside back cover.
The modulus of elasticity is one of the most important mechanical
properties used in the development of equations presented in this text. It
must always be remembered, though, that E, through the application of
Hooke's Jaw, Eq. 8- 3, can be used only if a material has linear elastic
behavior. Also, if the stress in the material is greater than the proportional
limit, the stress- strain diagram ceases to be a straight line, and so Hooke's
Jaw is no longer valid.
u
Strain Hardening. If a specimen of ductile material, such as steel, is
elastic
region
plastic
region
A'
0
0'
!permanent I elastic
set
recovery
(a)
u
elastic
region
0
O'
(b)
Fig. 8-14
plastic
region
loaded into the plastic region and then unloaded, elastic strain is recovered
as the material returns to its equilibrium state. The plastic strain remains,
however, and as a result the material will be subjected to a permanent set.
For example, a wire when bent (plastically) will spring back a little
(elastically) when the load is removed; however, it will not fully return to
its original position. This behavior is illustrated on the stress- strain diagram
shown in Fig. 8-14a. Here the specimen is loaded beyond its yield point A
to point A'. Since interatomic forces have to be overcome to elongate the
specimen elastically, then these same forces pull the atoms back together
when the load is removed, Fig. 8-14a. Consequently, the modulus of
elasticity, £ , is the same, and therefore the slope of line 0 'A' is the same as
line OA. With the load removed, the permanent set is 00 '.
If the load is reapplied, the atoms in the material will again be displaced
until yielding occurs at or near the stress A ', and the stress- strain
diagram continues along the same path as before, Fig. 8- 14b. Although
this new stress- strain diagram, defined by 0 'A ' B , now has a higher yield
point (A') , a consequence of strain hardening, it also has less ductility,
or a smaller plastic region, than when it was in its original state.
This pin was made of a hardened stee l
alloy, that is, one having a high carbon
content. It failed due to brittle fracture.
8.4
8.4
STRAIN ENERGY
STRAIN ENERGY
As a material is deformed by an external load, the load will do external work,
which in turn will be stored in the material as internal energy. This energy is
related to the strains in the material, and so it is referred to as strain energy.
To show how to calculate strain energy, consider a small volume element of
material take n from a tension test specimen, Fig. 8-lSa. It is subjected to the
uniaxial stress u.This stress develops a force 6.F = u 6A = u ( 6.x 6.y) on
the top and bottom faces of the element, which causes the element to
undergo a vertical displacement e 6.z , Fig. 8-lSb. By definition, work is
determined by the product of a force and displacement in the direction of
the force. He re the force is increased uniformly from zero to its final
magnitude 6.F when the displacement e 6.z occurs, and so during the
displacement the work done on the element by the force is equal to the
average force magnitude ( t.F/2) times the displacement e 6. z. The
conservation of energy requires this "e:i...-ternal work" on the element to be
equivalent to the "inte rnal work" or strain e nergy stored in the element,
assuming that no e nergy is lost in the form of heat. Consequently, the strain
energy is 6.U = ( t t::..F) E 6. z = (~ u 6.x 6.y) e 6.z. Since the vohume of
the element is 6. V = 6.x 6.y 6. z, then 6. U = ~ ae 6. V.
For e ngineering applications, it is often convenient to specify the strain
e nergy per unit volume of mate rial. This is called the strain energy
density , and it can be expressed as
6.U
6.V
LI= -
1
= -ae
2
Ay
(T
(a)
AF= <r(Ax Ay)
t
Ay
(8-6)
Finally, if the material behavior is linear elastic, then H ooke's law
applies, u = Ee, and therefore we can express the elastic strain energy
density in terms of the uniaxial stress u as
l u2
LI= - -
2 E
!
Free-body diagram
(b)
Fig. 8-15
(8-7)
(T
Modulus of R s e ce. When the stress in a material reaches the
proportional Limjt, the strain energy density, as calculated by Eq. 8-6 or
8-7, is referred to as the m odulus of resilience. It is
(8-8)
H ere u, is equivalent to the shaded triangular area under the elastic
region of the stress-strain diagram, Fig. 8-16a. Physically the modulus of
resilience represents the la rgest amount of strain energy per unit volume
the material can absorb without causing any permanent damage to the
material. Certainly this property becomes important when designing
bumpers or shock absorbers.
Ept
Modulus of resilience 11,
(a)
Fig. 8-16
389
390
CHAPTER
8
MECHAN IC AL PROPERTIES OF MATERIALS
Modulus of Toughness. Another important property of a material is
11,
Modulus of toughness 111
(b)
Fig. 8-16 (cont.)
hard steel
(0.6% carbon)
highest strength
structural steel
(0.2% carbon)
_ __.___ toughest
soft steel
(0.1 % carbon)
most ductile
its modulus of toughness, u 1• This quantity represents the entire area under
the stress- strain diagram, Fig. 8-16b,and therefore it indicates the maximum
amount of strain energy per unit volume the material can absorb just before
it fractures. Certainly this becomes important when designing members
that may be accidentally overloaded. By alloying metals, engineers can
change their resilience and toughness. For example, by changing
the percentage of carbon in steel, the resulting stress- strain diagrams in
Fig. 8-17 show how its resilience and toughness can be changed.
IMPORTANT POINTS
• A conventional stress- strain diagram is important in engineering
since it provides a means for obtaining data about a material's
tensile or compressive strength without regard for the material's
physical size or shape.
• Engineering stress and strain are calculated using the original
cross-sectional area and gage lengt h of the specimen.
• A ductile material, such as mild steel, has four distinct behaviors
as it is loaded. They are elastic behavior, yielding, strain
hardening, and necking.
• A material is linear elastic if the stress is proportional to the strain
within the elastic region. This behavior is described by Hooke's law,
a = Ee, where the modulus ofelasticity Eis the slope of the line.
Fi.g. 8-17
• Important points on the stress- strain diagram are the proportional
limit, elastic limit, yield stress, ultimate stress, and fracture stress.
• The ductility of a material can be specified by the specimen's
percent elongation or the percent reduction in area.
• If a material does not have a distinct yield point, a yield strength
can be specified using a graphical procedure such as the offset
method.
• Brittle materials, such as gray cast iron, have very little or no
yielding and so t hey can fracture suddenly.
• Strain hardening is used to establish a higher yield point for a
material. This is done by straining the material beyond the
elastic limit, then releasing the load. The modulus of elasticity
remains the same; however, the material's ductility decreases .
•
,
/
.
•
This nylon specimen exhibits a high
degree of toughness as noted by the large
amount of necking that has occurred just
before fracture.
• Strain energy is energy stored in a material due to its
deformation. This energy per unit volume is called strain
energy density. If it is measured up to t he proportional limit, it
is referred to as the modulus of resilience, and if it is measured
up to the point of fracture, it is called the modulus oftoughness.
It can be determined from t he area under the a-e diagram.
8.4
EXAMPLE
391
STRAIN ENERGY
8 .1
A tension test for a steel alloy results in the stress- strain diagram shown
in Fig. 8-18. Calculate the modulus of elasticity and the yield strength
based on a 0.2°10 offset. Identify on the graph the ultimate stress and the
fracture stress.
u (ksi)
120
u. = 108
Uys
=
o-~--­
-~1~10~------::;:::;-=---B
100
"' = 90 1 - - -,/-....- - -- - - -- - -__::,,, c
80
A'
70
68
60
50
40
30
20
10
0
/
,
,~
,
,
,
,,
,
,
,,
,
u
e1 = 0.23
IL--'--'---'--.l-,...i..:,----'--L--'-----'---'--'--"--'--
0.02 0.04 0.060.080.100.12 0.14 0.160.180.200.220.24
e (in. Jin.)
I o.OOos I o.0016 I o.obu
01%
0.0004 0.0012 0.0020
Fig. 8-18
SOLUTION
Modulus of Elasticity. We must calculate the slope of the initial
straight-line portion of the graph. Using the magnified curve and scale
shown in green, this line extends from point 0 to point A , which has
coordinates of approximately (0.0016 in.f in., 50 ksi). Therefore,
50 ksi
.
Ans.
E = 0.0016 in.f in. = 31.2{1CP) kst
Note that the equation of line OA is thusu = 31.2(103 )e.
Yield Strength. For a 0.2°10 offset, we begin at a strain of 0.2% or
0.0020 in.fin. and graphically extend a (dashed) line parallel to OA until
it intersects the u-E curve at A'. The yield strength is approximately
uys
=
68 ksi
Ans.
Ultimate Stress. This is defined by the peak of the u -e graph, point B
in Fig. 8-18.
u,, = 108 ksi
Ans.
Fracture Stress. When the specimen is strained to its maximum of
e1 = 0.23 in.fin., it (ractures at point C. Thus,
u1 = 90 ksi
Ans.
392
I
CHAPTER
EXAMPLE
8
MECHAN I CAL PROPERTIES OF MATERIALS
8.2
The stress- strain diagram for an aluminum alloy that is used for making
aircraft parts is shown in Fig. 8-19. If a specimen of this material is stressed
to a = 600 MPa, determine the permanent set that remains in the specimen
when the load is released. Also, find the modulus of resilience both before
and after the load application.
u (MPa)
750
600
uy= 450
;.\
parallel
300
150
G C
D
0 ~ 0.01 0.02 0.03
Ey = 0.006
0.023
~+-~~~---~-~- .- (mm/mm)
0.04
Eoc; -
Fig. 8-19
SOLUTION
When the specimen is subjected to the load, it
strain hardens until point Bis reached on the a -e diagram. The strain at
this point is approximately 0.023 mm/mm. When the load is released, the
material behaves by following the straight line BC, which is parallel to
line OA. Since both of these lines have the same slope, the strain at point
C can be determined analytically. The slope of line OA is the modulus of
elasticity, i.e.,
Permanent Strain.
450MPa
E = - - - - -= 75.0 GPa
0 .006 mm1mm
8.4
From triangle CBD, we require
BD
E = CD;
600(10 6 ) Pa
Pa CD
9)
_
75.0 ( 10
CD
=
0.008 mm/mm
This strain represents the amount of recovered elastic strain. The
permanent set or strain, e0 c, is thus
e0 c
=
0.023 mm/mm - 0.008 mm/mm
=
0.0150 mm/mm
Ans.
NOTE: If gage marks on the specimen were originally 50 mm apart, then
after the load is released these
( 0.0150) ( 50 mm) = 50.75 mm apart.
Modulus of Resilience.
marks
will
be
50 mm +
Applying Eq. 8- 8, the areas under OAG and
CBD in Fig. 8- 19 are*
(u,);nitial =
( u,) final
1
2
<TptEpt
1
2(450MPa) (0.006mm/mm)
=
=
1.35 MJ/m3
=
1
2
=
2.40 MJ/m3
<Tpt Ept
=
Ans.
1
( 600 MPa ) ( 0.008 mm/mm)
2
Ans.
NOTE: By comparison, the effect of strain hardening the material has
caused an increase in the modulus of resilience; however, note that
the modulus of toughness for the material has decreased, since the
area under the original curve, OABF, is larger than the area under
curve CBF.
*Work in the SJ system of units is measured in joules, where 1 J = 1 N · m.
STRAIN ENERGY
39 3
394
CHAPTER
8
MECHA NICAL PROPERTIES OF MATERIALS
FUNDAMENTAL PROBLEMS
F8-1. Define a homogeneous material.
F8-2. Indicate the points on the stress-strain diagram
which represent the proportional limit and the ultimate
stress.
F8-10. The material for the SO-mm-long specimen has the
stress-strain diagram shown. If P = 100 kN, determine the
elongation of the specimen.
F8-11. The material for the SO-mm-long specimen has the
stress-strain diagram shown. If P = ISO kN is applied and
then released, determine the permanent elongation of the
specimen.
u
D
p
u(MPa)
E
500 1-------=..,
450
~-~-
Prob.F8-2
~~~----~--
0.00225
F8-3. Define the modulus of elasticity E.
e (mm/mm)
0.03
Probs. FS-10/11
F8-4. At room temperature, mild steel is a ductile
material. True or false?
F8-S. Engineering stress and strain are calculated using
the actual cross-sectional area and length of the specimen.
True or false?
F8-6. As the temperature increases the modulus of
elasticity will increase. True or false?
F8-12. If t he elongation of wire BC is 0.2 mm after the
force P is applied, determine the magnitude of P. The wire
is A-36 steel and has a diameter of 3 mm.
F8-7. A 100-mm-long rod has a diameter of lS mm. If an
axial tensile load of 100 kN is applied, determine its change in
length. Assume linear elastic behavior with E = 200 GPa.
_ •gfc
F8-8. A bar has a length of 8 in. and cross-sectional area
of 12 in2• Determine the modulus of elasticity of the material
if it is subjected to an axial tensile load of 10 kip and
stretches 0.003 in. The material has linear elastic behavior.
F8-9. A IO-mm-diameter rod has a modulus of elasticity
of E = 100 GPa. If it is 4 m long and subjected to an axial
tensile load of 6 kN, determine its elongation. Assume
linear elastic behavior.
I
300mm
p
.._zoo mm'
.
B
, _ _ 400mm
I
Prob.F8-12
8.4
STRAIN ENERGY
39 5
PROBLEMS
8-L A tension test was performed on a steel specimen
having an original diameter of 0.503 in. and gage length of
2.00 in. The data is listed in the table. Plot the stress-strain
diagram and determine approximately the modulus of
elasticity. the yield stress, the ultimate stress., and the fracture
stress. Use a scale of I in. = 20 ksi and 1 in. = 0.05 in.fin.
Redraw the elastic region. using th e same stress scale but a
strain scale of l in.= 0.001 in.fin.
*8-4. The stress-strain diagram for a steel alloy having an
original diameter of0.5 in. and a gage length of2 in. is given in
the figure. Determine approximately the modulus of elasticity
for the material. the load on the specimen that causes yielding,
and the ultimate load the specimen will support.
u (ksi)
--
80
70
Load (kip) Elongation (in.)
0
1.50
0
0.0005
0.0015
0.0025
0.0035
4.60
8.00
L1.00
L1.80
L1.80
L2.00
0.0050
0.0080
0.0200
L6.60
0.0400
20.00
21.50
19.50
18.50
0.1000
0.2800
0.4000
0.4600
/
60
50
40
v
~
I
I
20
0
\
I
30
10
.........
I
0
/
0.04 0.08 0. L2 0.16 0.20 0.24 0.28
0 O.!Ul5
e (in.fin.)
o.rxn 0.1Xll5 O.IU2 O.IXJ2.~ O.IU3().ll l35
Prob. 8-4
Prob. 8-1
8-2. Data taken Crom a stress-strain test for a ceramic are
given in the table. The curve is linear between the origin and
the first point. Plot the diagram, and determine the modulus
of elasticity and the modulus of resilience.
8-3. Data taken from a stress~train test for a ceramic are
given in the table. The curve is linear between the origin and
the first point. Plot the diagram, and determine approximately
the modulus of toughness. The fracture stress is u1 = 53.4 ksi.
8-5. The stress-strain diagram for a steel alloy having an
original diameter of 0.5 in. and a gage length of 2 in. is given in
the figure. Uthe specimen is loaded until it is stressed to 70 ksi.
determine the approximate amount of elastic recovery and
the increase in the gage length after it is unloaded.
u {ksi)
80
e (in.f in.)
40
0
0.0006
0.()() I 0
0.0014
0.0018
0.0022
Probs. 8-213
0
~
/
I
I
20
10
\
I
30
0
33.2
45.5
49.4
51.5
53.4
..........
/
60
50
u (ks i)
-~
70
I
0
J
0.04 0.08 0. 12 0.16 0.20 0.24 0.28
0 0.(Xll5 O.IXJI 0.1Xll5 0.1Xl2 o.m25 O.IXJ"l.IU35
Prob. 8-5
e (in .fin.)
396
CHAPTER
8
MECHA NICAL PROPERTIES OF MATERIALS
8-6. The stress-strain diagram for a steel alloy having an
original diameter of 0.5 in. and a gage length of 2 in. is given
in the figure. Determine approximately the modulus of
resilience and the modulus of toughness for the material.
u (ksi)
80
CT
_,,.....
70
40
(psi)
b
/
'--- - - - - - - - - - - - E
I
I
20
0
r\
;f
30
10
-......
/
60
50
8-9. Acetal plastic has a stress-strain diagram as shown. If
a bar of this material has a length of 3 ft and cross-sectional
area of 0.875 in2 , and is subjected to an axial load of 2.5 kip,
determine its elongation.
I
(in.fin.)
Prob.8-9
J
o o.~ o.os 0.12 o.t6 0.20 o.24 o.28
0 0.0005 O.<Xll 0.0015 0.002 0.0025 O.<Xl30.0035
E
(in.fin.)
Prob.8-6
8-7. The rigid beam is supported by a pin at C and an A-36
steel guy wire AB. If the wire has a diameter of 0.2 in.,
determine how much it stretches when a distributed load of
w = 100 lb /ft acts on the beam. The material remains elastic.
*8-8. The rigid beam is supported by a pin at C and an
A-36 steel guy wire AB. If the wire has a diameter of 0.2 in.,
determine the distributed load w if the end B is displaced
0.75 in. downward.
8-10. The stress-strain diagram for an aluminum alloy
specimen having an original diameter of 0.5 in. and a gage
length of2 in. is given in the figure. Determine approximately
the modulus of elasticity for the material, the load on the
specimen that causes yielding, and the ultimate load the
specimen will support.
8-11. The stress-strain diagram for an aluminum alloy
specimen having an original diameter of 0.5 in. and a gage
length of2 in. is given in the figure. If the specimen is loaded
until it is stressed to 60 ksi, determine the approximate
amount of elastic recovery and the increase in the gage
length after it is unloaded.
*8-12. The stress-strain diagram for an aluminum alloy
specimen having an original diameter of 0.5 in. and a gage
length of2 in. is given in the figure. Determine approximately
the modulus of resilience and the modulus of toughness for
the material.
A
u (ksi)
70
60
50
~
40
30
20
•
w
1~----~ lOft ~------1
Probs. 8-7/8
10
,._..,...
I
I
-
.
- ~
-
I
'
I
O0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.2
0 0.0025 0.0050.0075 0.01 0.01250.0150.0175 0.02 0.0255 0.025
Probs. 8-10111112
E
(in.fin.)
8.4
8-13. A bar having a length of 5 in. and cross-sectional
area of 0. 7 in.2 is subjected to an axial force of 8000 lb. If the
bar stretches 0.002 in., determine the modulus of elasticity
of the material. The material has linear elastic behavior.
~..._________) ~ lb
i - - - - - 5 in.-- - - .
Prob.8-13
397
STRAIN ENERGY
8-17. The rigid beam is supported by a pin at C and an
A992 steel guy wire AB of length 6 ft. If the wire has a
diameter of 0.2 in., determine how much it stretches when a
distributed load of IV= 200 lb/ft acts on the beam. The wire
remains elastic.
8-18. The rigid beam is supported by a pin at C and an A992
steel guy wire AB of length 6 ft. If the wire has a diameter of
02 in., determine the distributed load IV if the end B is
displaced 0.12 in. downward. The wire remains elastic.
8-14. The rigid pipe is supported by a pin at A and an A-36
steel guy wire BO. If the wire has a diameter of 0.25 in.,
determine how much it stretches when a load of P = 600 lb
acts on the pipe.
8-15. The rigid pipe is supported by a pin at A and an A-36
guy wire BO. H the wire has a diameter of 0.25 in., determine
the load P if the end C is displaced 0.15 in. downward.
A
c
I
IV
l-------------100--------l
- - - 3 ft - - - t -- - 3 ft
-l
Probs. 8-17/18
Probs. 8-14115
*8-16. Direct tension indicators are sometimes used
instead of torque wrenches to ensure that a bolt has a
prescribed tension when used for connections. If a nut on
the bolt is tightened so that the six 3-mm high heads of the
indicator are strained 0.1 mm / mm, and leave a contact area
on each head of 1.5 mm2 , determine the tension in the bolt
shank. The material has the stress-strain diagram shown.
8-19. The stress-strain diagram for a bone is shown, and
can be described by the equation e = 0.45{1o-6) u +
0.36{ 10-12 ) u3, where u is in kPa. Determine the yield
strength assuming a 0.3% offset.
u (MPa)
600i-----~
u
4501--r-
1
ll
.r
- - -- - - - e (mm/ mm)
0.0015
0.3
Prob. 8-16
~------------€
Prob. 8-19
398
CHAPTER
8
MECHAN I CAL PROPERTIES OF MATERIALS
8.5
POISSON'S RATIO
When a deformable body is subjected to a force, not only does it elongate
but it also contracts laterally. For example, consider the bar in Fig. 8-20
that has an original radius r and length L , and is subjected to the tensile
force P. This force elongates the bar by an amount 8, and its radius
contracts by an amount 8' .The strains in the longitudinal or axial direction
and in the lateral or radial direction become
and
In the early 1800s, the French scientist S. D. Poisson realized that within
the elastic range the ratio of these strains is a constant, since the
displacements 8 and 8' are proportional to the same applied force. This
ratio is referred to as Poisson's ratio, v (nu), and it has a numerical value
that is unique for any material that is both homogeneous and isotropic.
Stated mathematically it is
(8- 9)
When the rubber block is
compressed (negative sLrain), its
sides will expand (positive strain).
The ratio of these strains remains
constant.
The negative sign is included here since longitudinal elongation (positive
strain) causes lateral contraction (negative strain), and vice versa. Keep
in mind that these strains are caused only by the single axial or
longitudinal force P; i.e., no force acts in a lateral direction in order to
strain the material in this direction.
Poisson's ratio is a dimensionless quantity, and it will be shown in
Sec. 10.6 that its maximum possible value is 0.5, so that 0 ~ v ~ 0.5. For
most nonporous solids it has a value that is generally between 0.25 and
0.355. Typical values for common engineering materials are listed on the
inside back cover.
Tension
Fig. 8-20
8.5
I EXAMPLE
8.3
A bar made of A-36 steel has the dimensions shown in Fig. 8-21. If an axial
force of P = 80 kN is applied to the bar, determine the change in its length
and the change in the dimensions of its cross section. The material behaves
elastically.
."
P=80kN
y
SO mm
x
lOOmm
~"--z
Fig. 8-21
SOLUTION
The normal stress in the bar is
N
az
=A=
80(103 ) N
(0.1 m) (0.05 m)
=
16.0(106) Pa
From the table on the inside back cover for A-36 steel
and so the strain in the z direction is
az
e = =
z
Est
16.0(106 ) Pa
200(109 ) Pa
=
Est =
200 GPa,
80(10-·{i) mm/mm
The axial elongation of the bar is therefore
Ans.
Using Eq. 8-9, where vst = 0.32 as found from the inside back cover, the
lateral contraction strains in both the x and y directions are
Thus the changes in the dimensions of the cross section are
Bx
= ExLx =
By = ey L y =
-(25.6(10-6))(0.1 m) = -2.56µm
-(25.6(10-6)](0.05m) = -1.28µm
Ans.
Ans.
POISSON'S RATIO
399
400
CHAPTER
8
MECHAN I CAL PROPERT I ES OF MATERIALS
8.6
f
T zy
IO_f
In Sec. 7.5 it was shown that when a small element of material is subjected
to pure shear, equilibrium requires that equal shear stresses must be
developed on four faces of the element, Fig. 8-22a. Furthermore, if the
material is homogeneous and isotropic, then this shear stress will distort
the element uniformly, Fig. 8- 22b, producing shear strain.
In order to study the behavior of a material subjected to pure shear,
engineers use a specimen in the shape of a thin tube and subject it to a
torsional loading. If measurements are made of the applied torque and
the resulting angle of twist, then by the methods to be explained in
Chapter 10, the data can be used to determine the shear stress and shear
strain within the tube and thereby produce a shear stress- strain diagram
such as shown in Fig. 8- 23. Like the tension test, this material when
subjected to shear will exhibit linear elastic behavior and it will have a
defined proportional limil Tp/· Also, strain hardening will occur until an
ultimate shear stress Tu is reached. And finally, the material will begin to
Jose its shear strength until it reaches a point where it fractures, T1.
For most engineering materials, like the one just described, the elastic
behavior is linear, and so Hooke's Jaw for shear can be written as
x
(a)
y
7T -
2
THE SHEAR STRESS-STRAIN
DIAGRAM
'Y:rv
·.
(b)
Fi.g. 8-22
(8-10)
......-
'
-
-.........
/
G
"Ip/
y,,
Yr
Here G is called the shear modulus of elasticity or the modulus of
rigidity. Its value represents the slope of the line on the T- y diagram,
that is, G = Tpi/'Ypl· Units of measurement for G will be the same as those
for T (Pa or psi), since y is measured in radians, a dimensionless quantity.
Typical values for common engineering materials are listed on the inside
back cover.
Later it will be shown in Sec. 14.11 that the three material constants,
E, v, and G can all be related by the equation
Fig. 8-23
(8-11)
Therefore, if E and Gare known, the value of v can then be determined
from this equation rather than through experimental measurement.
8.6
THE SHEAR STRESS-STRAIN D IAGRAM
401
A specimen of titanium alloy is tested in torsion and the shear T (ksi)
stress-strain diagra m is shown in Fig. 8-24a. Determine the shear 90
modulus G, the proportional limit, and the ultimate shear stress. Also, 80
B
dete rmine the maximum distance d that the top of a block of this ~tT.;;-3!:;;:::::;:=....--T°'-.....
mate rial, shown in Fig. 8-24b. could be displaced horizontally if the 50
mate rial behaves elastically when acted upon by a shear force V. What is
the magnitude of V necessary to cause this displacement?
20
;g
10
SOLUTION
Oyp1
Shear Modulus. This value represents the slope of the straight-line
portion OA of the r - y diagram. The coordinates of point A are (0.008 rad,
52 ksi). Thus,
G =
52 ksi
_ rad
0 008
The equation of line OA is therefore,,.
law for shear.
Proportional Limit.
point A. Thus,
=
.
6500 ks1
=
Cy
=
E
(b}
Fig. 8-24
T 11
= 52 ksi
Ans.
= 73 ksi
Ans.
Maximum Elastic Displacement and Shear Force. The shear strain
at the comer C of the block in Fig. 8-24b is determined by finding the
difference in the 90° angle DCE and the angle 8. This angle is y = 90° - 8
as shown. From the r-y diagram the maximum elastic shear strain is
0.008 rad, a very small angle. The top of the block in Fig. 8-24b will
therefore be displaced horizontally a distanced given by
tan ( 0.008 rad ) ""' 0.008 rad
d
= -
d
.
2 m.
= 0.016 in.
The corresponding average shear stress in the block is r,,1
Thus, the shear force V needed to cause the displacement is
~;
(a)
6500y, which is Hooke's
Ultimate Stress. This value represents the maximum shear stress,
point B. From the graph,
=
y. = 0.54 0.73 Y (rad}
Ans.
By inspection, the graph ceases to be linear at
Tp/
7'avg
= 0.008
52 ksi
=
52 ksi.
= (3 in.~( 4 in.)
v = 624 kip
Ans.
402
I
CHAPTER
EXAMPLE
8
MECHAN IC AL PROPERTIES OF MATERIALS
8.5
An aluminum specimen shown in Fig. 8-25 has a diameter of d 0 = 25 mm
and a gage length of L 0 = 250 mm. If a force of 165 kN elongates the gage
length 1.20 mm, determine the modulus of elasticity. Also, determine by how
much the force causes the diameter of the specimen to contract. Take
G.1 = 26 GPa and ay = 440 MPa.
165 kN
t
f
SOLUTION
Modulus of Elasticity. The average normal st ress in the specimen is
P
a = Lo
165(1a3) N
=
A
=
(7r/4)(0.025m) 2
336.1 MPa
and the average normal strain is
e = -
5
L
1.20 mm
250mm
=
=
0.00480 mm/mm
Since a < ay = 440 MPa, the material behaves elastically. The modulus
of elasticity is therefore
a
i
165 kN
Fig. 8-25
E a1 =
-;- =
336.1(106 ) Pa
0.004 0
8
=
70.0 GPa
Ans.
Contraction of Diameter. Fi1rst we will determine Poisson's ratio for
the material using Eq. 8-11.
E
G =---2(1 + v)
26GPa
70.0 GPa
2(1 + v)
= ----
v = 0.347
Since EJong
=
0.00480 mm/mm, then by Eq. 8-9,
v =
0.347
EJat
=
=
0.00480 mm/mm
-0.00166 mm/mm
The contraction of the diameter is therefore
5' = (0.00166) (25 mm)
=
0.0416mm
Ans.
8.6
403
THE SHEAR STRESS-STRAIN DIAGRAM
IMPORTANT POINTS
• Poisson's ratio, 11,is a ratio of the lateral strain of a homogeneous
and isotropic material to its longitudinal strain. Generally
these strains are of opposite signs, that is, if one is an elongation,
the other will be a contraction.
• The shear stress-strain diagram is a plot of the shear stress
versus the shear strain. If the material is homogeneous and
isotropic, and is also linear elastic, the slope of the straight line
within the elastic region is called the modulus of rigidity or the
shear modulus, G.
• There is a mathematical relationship between G, E, and v.
FUNDAMENTAL PROBLEMS
FS-13. A 100 mm long rod has a diameter of 15 mm. If an
axial tensile load of 10 kN is applied to it, determine the
0.35.
change in its diameter. E = 70 GPa, 11
=
FS-15. A 20-mm-wide block is firmly bonded to rigid
plates at its top and bottom. When the force P is applied the
block deforms into the shape shown by the dashed line.
Determine the magnitude of P.1l1e block"s material has a
modulus of rigidity of G 26 G Pa. Assume that the material
does not yield and use small angle analysis.
=
J-150mrn--j
0.5 mml-t
I
• l-t... A solid circular rod that is 600 mm long and 20 mm
in diameter is subjected to an axial force of P = 50 kN The
I
elongation of the rod is Ii = l.40 mm. and its diameter
becomes d ' = 19.9837 mm. Determine the modulus of
elasticity and the modulus of rigidity of the material.
Assume that the material does not yield.
150mm
p
.
I
I
I
I
'
I
I
1
I
I
I
I
•
p 1b
..... _
H~-16.
P = 50kN
A 20.mm-wide block is bonded to rigid plates at its
top and bottom. When the force P is applied the block deforms
into the shape shown by the dashed line. U a= 3 mm and P is
released, determine the permanent shear strain in the block.
~600mm
T
l
no ~-~----
P=50kN
Prob. FS-14
,-----.
,
r
-150 nun-J
a.1= 3 mm
I
(MPa)
I
~,...,..,,,,.....---y
(rad)
0.005
150mm :
L1
Proh.I-S-i6
I
/
,'
p
404
CHAPTER
8
MECHA NICAL PROPERTIES OF MATERIALS
PROBLEMS
*8-20. The acrylic plastic rod is 200 mm long and 15 mm in
diameter. If an axial load of 300 N is applied to it, determine
the change in its length and the change in its diameter.
EP = 2.70 GPa, "P = 0.4.
~~---==================-~+--...
DN
IDN
8-22. The elastic portion of the stress-strain diagram for
an aluminum alloy is shown in the figure. The specimen
from which. it was obtained has an original diameter of
12.7 mm and a gage length of 50.8 mm. When the applied
load on the specimen is 50 kN, the diameter is 12.67494 mm.
Determine Poisson's ratio for the material.
8-23. The elastic portion of the stress-strain diagram for an
aluminum alloy is shown in the figure. The specimen from
which it was obtained has an original diameter of 12.7 mm
and a gage length of 50.8 mm. If a load of P = 60 kN is
applied to t he specimen, determine its new diameter and
length. Take " = 0.35.
1 - - - - - - 200 mm - - - - - - 1.
u (MPa)
Prob. 8-20
8-2L The plug has a diameter of 30 mm and fits within a
rigid sleeve having an inner diameter of 32 mm. Both the
plug and the sleeve are 50 mm long. Determine the axial
pressure p that must be applied to the top of the plug to
cause it to contact the sides of the sleeve. Also, how far must
the plug be compressed downward in order to do this? The
plug is made from a material for which E = 5 MPa," = 0.45.
Probs. 8-22/23
*8-24. The brake pads for a bicycle tire are made of
rubber. If a frictional force of 50 N is applied to each side of
the tires, determine the average shear strain in the rubber.
Each pad has cross-sectional dimensions of 20 mm and
50 mm. G, = 0.20 MPa.
SO mm
Prob. 8-21
Prob.8-24
8.6
8-25. The lap joint is connected together using a 1.25 in.
diameter bolt. U the bolt is made from a material having a
shear streslH!train diagram that is approximated as shown,
determine the shear strain developed in the shear plane of
the bolt when P = 75 kip.
THE SH EAR STRESS-STRAIN D IAGRAM
405
*8-28. The shear strcslH!train diagram for an alloy is
shown in the figure. U a bolt having a diameter of 0.25 in. is
made of this material and used in the lap joint, determine
the modulus of elasticity E and the force P required to
cause the material to yield. Take "= 0.3.
8-26. The lap joint is connected together using a 125 in.
diameter bolt. U the bolt is made from a material having a
shear streslH!train diagram that is approximated as shown,
determine the permanent shear strain in the shear plane of
the bolt when the applied force P = 150 kip is removed.
T
Ty=
(ksi)
50
f
0.004
p
2
T
(ksi)
Prob. 8-28
< - - - J . - - - -- - - ' --
0.005
Y (rad)
0.05
Probs. 8-25/26
8-29. A shear spring is made from two blocks of rubber,
each having a height h, width b. and thickness a. The blocks
are bonded to three plates as shown. If the plates are rigid
and the shear modulus of the rubber is G, determine the
displacement of plate A when the vertical load P is applied.
Assume that the displacement is small so that
li = a tan 1' = ay.
8-27. The rubber block is subjected to an elongation of
0.03 in. along the x axis, and its vertical faces are given a tilt
so that (J = 89.3°. Determine the strains Ex• E1 and1'.ry· Take
,,, = 0.5.
p
--. 111T
y
I
.
3 Lil.
r·-------------------- ---;
,
,
,
LJ ----------------------- _f:J_
1---1
4 in.
I
- -- - 1
1
Prob. 8-27
J"
I •
x
_... ... ----a- -a-
rI
Prob.8-29
406
C H APT ER
8
M ECHANICAL PROPERT IES OF MATERIALS
CHAPTER REVIEW
One of the most important tests for material strength is the tension test. The results, found from
stretching a specimen of known size, are plotted as normal stress on the vertical axis and normal
strain on the horizontal axis.
u
Many engineering materials exhibit
initial linear elastic behavior, whereby
stress is proportional to strain, defined
by Hooke's Jaw, u = EtE. Here E, called
the modulus of elasticity, is the slope of
this straight line on the stress-strain
diagram.
u
= EtE
E u
IE
IE
Ductile material
When the material is stressed beyond
the yield point, permanent deformation
will occur. In particular, steel has a
region of yielding, whereby the material
will exhibit an increase in strain with no
increase in stress. The region of strain
hardening causes further yielding of the
material with a corresponding increase
in stress. Finally, at the ultimate stress, a
localized region on the specimen will
begin to constrict, forming a neck. It is
after this that the fracture occurs.
Ductile materials, such as most metals,
exhibit both elastic and plastic behavior.
Wood is moderately ductile. Ductility is
usually specified by the percent
elongation to failure or by the percent
reduction in the cross-sectional area.
Brittle materials exhibit little or no
yielding before failure. Cast iron,
concrete, and glass are typical examples.
u
~ultimate
u,,
proportional limit ~
\ ~lastic limit
yield stress
Uf
Uy
Up/
I
~
E
elastic yielding
strain
necking
region
hardening
plastic behavior
elastic
behavior
Conventional and true stress-strain diagrams
for ductile material (steel) (not to scale)
Percent elongation
fracture
stress
=
Lr - Lo
Percent reduction of area
Lo
=
(100% )
Ao - A
Ao f ( 100% )
CHAPTER REVIEW
The yield point of a material at A can
be increased by strain hardening. This
is accomplished by applying a load that
causes the stress to be greater than the
yield stress, then releasing the load.
The larger stress A' becomes the new
yield point for the material.
(]"
Poisson's ratio v is a dimensionless
material property that relates the
lateral strain to the longitudinal strain.
Its range of values is 0 < v < 0.5.
plastic
region
elastic
region
set
When a load is applied to a member,
the deformations cause strain energy to
be stored in the material. The strain
energy per unit volume or strain energy
density is equivalent to the area under
the stress-strain curve. This area up to
the yield point is called the modulus of
resilience. The entire area under the
stress-strain diagram is called the
modulus of toughness.
elastic
recovery
(]"
O"p/ f - - - --f
11,
11,
Modulus of toughness 111
Ep/
(b)
Modulus of resilience
E1a1
p
v= - - Etong
Tension
Shear stress versus shear strain diagrams
can also be established for a material.
Within the elastic region, r = G-y,
where G is the shear modulus found
from the slope of the line. The value of v
can be obtained from the relationship
that exists between G , E, and v.
407
7
E
G= - - - 2(1 + v)
408
CHAPTER
8
MECHA NICAL PROPERTIES OF MATERIALS
REVIEW PROBLEMS
R8-1. The elastic portion of the tension stress-strain
diagram for an aluminum alloy is shown in the figure. The
specimen used for the test has a gage length of 2 in. and a
diameter of 0.5 in. When the applied load is 9 kip, the new
diameter of the specimen is 0.49935 in. Calculate the shear
modulus Ga1 for the aluminum.
R8-2. The elastic portion of the tension stress-strain
diagram for an aluminum alloy is shown in the figure. The
specimen used for the test has a gage length of 2 in. and a
diameter of 0.5 in. If the applied load is 10 kip, determine
the new diameter of the specimen. The shear modulus is
Ga1 = 3.8( lfrl) ksi.
*R8-4. The wires each have a diameter oft in., length of
2 ft, and are made from 304 stainless steel. If P = 6 kip,
determine the angle of tilt of the rigid beam AB.
R8-S. The wires each have a diameter of t in., length of
2 ft, and are made from 304 stainless steel. Determine the
magnitude of force p so that the rigid beam tilts 0.015°.
D
1
AB--2" f l < - l
2 ft
u (ksi)
I
-
-·
-
I
Probs. R8-4/S
"---- - - ' - - - - - - - - - E
0.00614
(in.fin.)
Probs. RS-112
R8-3. The rigid beam rests in the horizontal position on
two 2014-T6 aluminum cylinders having the unloaded
lengths shown. If each cylinder has a diameter of 30 mm,
determine the placement x of the applied 80-kN load so
that the beam remains horizontal. What is the new diameter
of cylinder A after the load is applied? v31 =0.35.
R8-6. The head H is connected to the cylinder of a
compressor using six 136 in. diameter steel bolts. If the
clamping force in each bolt is 800 lb, determine the normal
strain in the bolts. If uy = 40 ksi and Esi = 29 ( 103 ) ksi,
what is the strain in each bolt when the nut is unscrewed so
that the clamping force is released?
CL
H
I
220 n1ml
1
8
IA
I l2fomm
1- - 3m - -1
Prob. R8-3
Prob. R8-6
409
REVIEW PROBLEMS
R8-7. The stress--strain diagram for polyethylene, which is
used to sheath coaxial cables, is determined from testing a
specimen that has a gage length of 10 in. If a load Pon the
specimen develops a strain of £ = 0.024 in./in., determine
the approximate length of the specimen, measured between
the gage points, when the load is removed. Assume the
specimen recovers elastically.
R8-9. The 8-mm-diamctcr bolt is made of an aluminum
alloy. It fits through a magnesium sleeve that has an inner
diameter of 12 mm and an outer diameter of 20 mm. If the
original lengths of the bolt and sleeve are 80 mm and
SO mm. respectively. determine the strains in the sleeve and
the bolt if the nut on the bolt is tightened so that the tension
in the bolt is 8 kN. Assume the material at A is rigid.
£a1 = 70 GPa, Emg = 45 GPa.
u (ksi)
p
5
3
I
2
1
I
..-<'
4
<
0
v
"0
50mm
I
p
f
00
e (in.fin.)
0.008 0.016 0.024 0.032 0.040 0.048
Prob. R8-7
Prob. R8-9
*R8-8. The pipe with two rigid caps anached to its ends is
subjected 10 an axial force P. If the pipe is made from a
material having a modulus of elasticity £ and Poisson's
ratio v. determine the change in volume of the material.
R8-10. An acetal polymer block is fixed to the rigid plates
at its top and boltom surfaces. If the top plate displaces
2 mm horizontally when it is subjected to a horizontal force
P = 2 kN, determine the shear modulus of the polymer. The
width of the block is 100 mm. Assume that the polymer is
linearly elastic and use small angle analysis.
T·
To
L
p
~Section a - a
1- - - 4 0 0 m m - - -1
P=2 kN
I
L
200mm
p
Prob. R8-8
Prob. R8-10
CHAPTER
8
(© Hazlan Abdul Hakim/Getty Images)
The string of drill pipe stacked on this oil rig will be subjected to large axial
deformations when it is placed in the hole.
AXIAL LOAD
CHAPTER OBJECTIVES
•
In this chapter we will discuss how to determine the deformation
of an axially loaded member, and we will also develop a method
for finding the support reactions when these reactions cannot be
determined strictly from the equations of equilibrium. An analysis
of the effects of thermal stress, stress concentrations, inelastic
deformations, and residual stress will also be discussed.
9.1
SAINT-VENANT'S PRINCIPLE
In the previous chapters, we have developed the concept of stress as a
means of measuring the force distribution within a body and strain as a
means of measuring a body's deformation. We have also shown that the
mathematical relationship between stress and strain depends on the type
of material from which the body is made. In particular, if the material
behaves in a linear elastic manner, then Hooke's Jaw applies, and there is
a proportional relationship between stress and strain.
Using this idea, consider the manner in which a rectangular bar will
deform elastically when the bar is subjected to the force P applied! along
its centroidal axis, Fig. 9- la. The once horizontal and vertical grid lines
drawn on the bar become distorted, and Localized deformation occurs at
each end. Throughout the midsection of the bar, the lines remain
horizontal and vertical.
411
412
CHAPTER
9
AXIAL L OAD
p
t :::;::~:=-
..- ~
a-
Load distorts lines
located near load
a
b-
b
c
. . . .: :t::=- Lines located away
from the load and support
remain straight
!-+-<-+-+
Load distorts lines
located near support
(a)
Fig. 9-1
If the material remains elastic, then the strains caused by this
deformation are directly related to the stress in the bar through Hooke's
law, a = Ee. As a result, a profile of the variation of the stress distribution
acting at sections a-a, b- b, and c- c, will look like that shown in Fig. 9-lb.
Notice how the lines on this rubber
membrane distor t after it is stretched. The
localized distortions at the grips smooth out
as stated by Saint-Ve nant's principle.
By comparison, the stress tends to reach a uniform value at section c-c,
which is sufficiently removed from the end since the localized
deformation caused by P vanishes. The minimum distance from the bar's
end where this occurs can be determined using a mathematical analysis
based on the theory of elasticity. It has been found that this distance
should at least be equal to the largest dimension of the loaded cross
section. Hence, section c-c should be located at a distance at least equal
to the width (not the thickness) of the bar.*
In the same way, the stress ,distribution at the support in Fig. 9-la will
also even out and become uniform over the cross section located the
same distance away from the support.
The fact that the localized stress and deformation behave in this
manner is referred to as Saint- Venant's principle, since it was first
noticed by the French scientist Barre de Saint-Venant in 1855. Essentially
it states that the stress and strain produced at points in a body sufficiently
removed from the region of external load application will be the same as
the stress and strain produced by any other applied external loading that
has the same statically equivalent resultant and is applied to the body
within the same region. For example, if two symmetrically applied forces
P /2 act on the bar, Fig. 9- lc, the stress distribution at section c-c will be
uniform and therefore equivalent to aavg = P /A as in Fig. 9-lc.
*When section c--c is so located, the theory of elasticity predicts the maximum stress to
be Umax = 1.02 Uavg·
9.2
ELASTIC DEFORMATION OF AN AxlALLY LOADED MEMBER
p
p
u.,,. =
section a-a
41 3
section l>-b
p
A
section c-c
u.,.
p
-a:
A
section c-c
(b)
(c)
Fig. 9-1 (cont.)
9.2
ELASTIC DEFORMATION OF AN
AXIALLY LOADED MEMBER
Using Hooke's law and the definitions of stress and strain, we will now
develop an equation that can be used to determine the elastic displacement
of a member subjected to axial loads. To generalize the development,
consider the bar shown in Fig. 9-2a, which has a cross-sectional area that
gradually varies along its length L, and is made of a material that has a
variable stiffness or modulus of elasticity. The bar is subjected to
concentrated loads at its ends and a variable external load distributed
along its length. This distributed load could, for example, represent the
weight of the bar if it is in the vertical position, or friction forces acting on
the bar's surface.
Here we wish to find the relative displacement 8 (delta) of one end of
the bar with respect to the other end as caused by the loading. We will
neglect the localized deformations that occur at points of concentrated
loading and where the cross section suddenly changes. From SaintVenant's principle, these effects occur within small regions of the bar's
length and will therefore have only a slight effect on the final result. For
the most part, the bar will deform uniformly, so the normal stress will be
uniformly distributed over the cross section.
t--x--11- dx
Pi
"'oe1-----1I- - - -
_,
!!- - - - ]1-L-.J-;..~
----L
(a)
Fig. 9-2
Isl
P2
TI1e vertical displacement of the rod at the
top floor B only depends upon the force in
the rod along length AB. However, the
displacement at the bottom floor C depends
upon the force in the rod along its entire
length, ABC.
414
CHAPTER
9
AXIAL L OAD
1----x---l l - dx
P,
.,.4---11- - - -
1----lt
II I
~~~~ L
(a)
N (x) •
dx-
l_D
m:
Using the method of sections, a differential element (or wafer) of
length dx and cross-sectional area A(x) is isolated from the bar at the
arbitrary position x, where the modulus of elasticity is E(x). The freebody diagram of this element is shown in Fig. 9- 2b. The resultant internal
axial force will be a function of x since the external distributed loading
will cause it to vary along the length of the bar. This load, N(x), will
deform the element into the shape indicated by the dashed outline, and
therefore the displacement of one end of the element with respect to the
other e~? becomes do. The stress and strain in the element are therefore
!I •
P2
..!
~
N(x)
A(x)
do
dx
<T = - - and € = -
Provided the stress does not exceed the proportional limit, we can apply
Hooke's Jaw; i.e., <T = E(x)e, and so
• N(x)
N(x)
A(x)
-1-i- dll
=
E( )(do)
x dx
1
(b)
do
Fig. 9-2 (Repeated)
=
N(x)dx
A(x)E(x)
For the entire length L of the bar, we must integrate this expression to
find 0. This yields
8-
1
L
0
N(x)dx
A(x)E(x)
(9- 1)
Here
o=
L =
N (x) =
A(x)
E(x)
=
=
displacement of one point on the bar relative to the other point
original length of bar
internal axial force at the section, located a distance x from
one end
cross-sectional area of the bar expressed as a function of x
modulus of elasticity for the material expressed as a function
ofx
Constant Load and Cross-Sectional Area. In many cases the
bar will have a constant cross-sectional area A; and the material will be
homogeneous, so E is constant. Furthermore, if a constant external force
is applied at each end, Fig. 9- 3a, then the internal force N throughout the
length of the bar is also constant. As a result, Eq. 9- 1 when integrated
becomes
~
~
(9- 2)
9.2
x
p
p
L
415
ELASTIC DEFORMATION OF AN AxlALLY LOADED MEMBER
P,
~
ll
~2
-L,~-Li-l
I
(a)
P3
I
• P4
~---l-L.-1
(b)
Fig.9-3
If the bar is subjected to several different axial forces along its length,
or the cross-sectional area or modulus of elasticity changes abruptly
from one region of the bar to the next, as in Fig. 9- 3b, then the above
equation can be applied to each segment ofthe bar where these quantities
remain constant. The displacement of one end of the bar with respect to
the other is then found from the algebraic addition of the relative
displacements of the ends of each segment. For this general case,
-I
E
I------~
+N
•
~I
+s
(9- 3)
Sign Convention. When applying Eqs. 9- 1 through 9- 3, it is best to
use a consistent sign convention for the internal axial force and the
displacement of the bar. To do so, we will consider both the force and
displacement to be positive if they cause tension and elongation, Fig. 9-4;
whereas a negative force and displacement will cause compression and
contraction.
IMPORTANT POINTS
• Saint-Venant'.s principle states that both the localized
deformation and stress which occur within the regions of load
application or at the supports tend to "even out" at a distance
sufficiently removed from these regions.
• The displacement of one end of an axially loaded member relative
to the other end is determined by relating the applied internal
load to the stress using <T = N /A and relating the displacement
to the strain using e = do/dx. Fmally these two equations are
combined using Hooke's Jaw, <T = Ee, which yields Eq. 9- 1.
• Since Hooke's Jaw has been used in the development of the
displacement equation, it is important that no internal load
causes yielding of the material, and that the material behaves
in a linear elastic manner.
+N
...
1]
l~'--------l
+s
Fi.g. 9-4
416
CHAPTER
9
AXIAL LOAD
PROCEDURE FOR ANALYSIS
The relative displacement between any two points A and B on an
axially loaded member can be determined by applying Eq. 9-1 (or
Eq. 9- 2). Application requires the following steps.
Internal Force.
• Use the method of sections to determine the internal axial
force N within the member.
• If this force varies along the member's length due to an external
distributed loading, a section should be made at the arbitrary
location x from one end of the member, and the internal force
represented as a function of x, i.e., N(x).
• If several constant external forces act on the member, the internal
force in each segment of the member between any two external
forces must be determined.
• For any segment, an internal tensile force is positive and an
internal compressive force is negative. For convenience, the
results of the internal loading throughout the member can be
shown graphically by constructing the normal-force diagram.
Displacement.
• When the member's cross-sectional area varies along its length,
the area must be expressed as a function of its position x ,
i.e.,A(x).
• If the cross-sectional area, the modulus of elasticity, or the
internal loading suddenly changes, then Eq. 9-2 should be
applied to each segment for which these quantities are
constant.
• When substituting the data into Eqs. 9- 1 through 9-3, be sure to
account for the proper sign of the internal force N. Tensile forces
are positive and compressive forces are negative. Also, use a
consistent set of units. For any segment, if the result is a positive
numerical quantity, it indicates elongation; if it is negative, it
indicates a contraction.
9.2
I
EXAMPLE
ELASTIC DEFORMATION OF AN AxlALLY LOADED MEMBER
417
9.1
The uniform A-36 steel bar in Fig. 9-5a has a diameter of 50 mm and is
subjected to the loading shown. Determine the displacement at D , and the
displacement of point B relative to C.
Nev= 70 kN
80kN
-il-1m-
8
1-A_ _
z m---
40 kN
Jlo tr"
. ~~l
70 kN
70kN
Nsc = 50 kN
l=--1.5 m--ID
.,r
___so_ k_N_ _40_k_N_--=;=J • 70 kN
(a)
(b)
Fig. 9-5
SOLUTION
The internal forces within the bar are determined
using the method of sections and horizontal equilibrium. The r,esults are
shown on the free-body diagrams in Fig. 9- 5b. The normal-force diagram
in Fig. 9- 5c shows the variation of these forces along the bar.
Internal Forces.
From the table on the inside back cover, for A-36 steel,
E = 200 GPa. Using the established sign convention, the displacement of
the end of the bar is therefore
N (kN)
Displacement.
L
NL
AE
=
[-70(1a3) N)(l.5 m)
?T(0.025 m) 2[200(109) N/m2]
50 1--- - - 1----~1--~3.,__ ____:,
4·;5
::..- x(m)
-30 +----- 2 L..-~~
- 70
[50(103) N](2 m)
(-30(HP) N](l m)
+ - - - -2- - -9- - + - - - -2- - -9- - 2
2
?T(0.025 m) [200(10 ) N /m
OD
=
]
?T(0.025 m) [200(10
)
-89.1(10- 3) mm
N /m
]
Ans.
This negative result indicates that point D moves to the left.
The displacement of B relative to C, 5810 is caused only by the internal
load within region BC. Thus,
f>s;c =
NL
AE
=
[-30(103) N)(l m)
?T(0.025 m) 2[200(109) N/m2]
=
_3
6
-? .4(lO ) mm
Here the negative result indicates that B will move towards C.
Ans.
(c)
418
I
CHAPTER
EXAMPLE
9
AXIAL L OAD
9.2
The assembly shown in Fig. 9--6a consists of an aluminum tube AB having a
cross-sectional area of 400 mm2 . A steel rod having a diameter of 10 mm is
attached to a rigid collar and passes through the tube. If a tensile load of
80 kN is applied to the rod, determine the displacement of the end C of the
rod. Take E st = 200 GPa, E al = 70 GPa.
~ 400mm
NAB= 80 kN
80 kN --41-~~-----i•---
(b)
(a)
Fig. 9-6
SOLUTION
The free-body diagrams of the tube and rod segments
in Fig. 9-6b show that the rod is subjected to a tension of 80 kN, and the
tube is subjected to a compression of 80 kN.
Internal Force.
We will first determine the displacement of C with respect
to B. Working in units of newtons and meters, we have
Displacement.
f>qs =
NL
AE
=
[+ 80(103) NJ (0.6 m)
?T(0.005 m) 2 [200 (109) N/m2]
=
+ 0.003056 m ~
The positive sign indicates that C moves to the right relative to B, since
the bar elongates.
The displacement of B with respect to the fixed end A is
_ NL
[-80(103) N](0.4 m)
08
- AE - (400 mm2 (10-6) m2/mm2][70(109) N/m2]
=
-0.001143 m
=
0.001143 m ~
Here the negative sign indicates that the tube shortens, and so B moves
to the right relative to A.
Since both displacements are to the right, the displacement of C
relative to the fixed end A is therefore
5c
=
5 8 + 5 C/B
=
0.00420 m
=
=
0.001143 m + 0.003056 m
4.20 mm ~
Ans.
9.2
EXAMPLE
419
ELASTIC DEFORMATION OF AN AxlALLY LOADED MEMBER
9 .3
Rigid beam AB rests on the two short posts shown in Fig. 9-?a.AC is made
of steel and has a diameter of20 mm, and BD is made of aluminum and has
a diameter of 40 mm. Determine the displacement of point Fon AB if a
vertical load of 90 kN is applied over this point. Take E,1 = 200 GPa,
E;,1 = 70 GPa.
90kN
1200 mmi -400 mm-J
A
-,
8
F
300mm
SOLUTION
The compressive forces acting at the top of each post
are determined from the equilibrium of member AB, Fig. 9-7b. These
forces are equal to the internal forces in each post, Fig. 9-7c.
Internal Force.
Displacement.
vi
c
(a)
90kN
The displacement of the top of each post is
200 mm
---.t
1
Post AC:
400 mm
--1
I
t
60kN
1
30kN
(b)
= 0.286 mm!
60kN
30kN
Post BD:
S8
-
N 80 L 80
[-30{HY)N]{0.300m)
7T{0.020 m)2[70{109) N/m2]
-
AsoE.1
=
t
= 0.102 mm!
A diagram showing the centerline displacements at A, B, and Fon the
beam is shown in Fig. 9-7d. By proportion of the blue shaded triangle,
the displacement of point Fis therefore
SF=
0.102 mm + (0.184 mm)(
I
0.102mm A
400
mm)
600mm
600mm
=
0.225 mm!
I
i- - 400mm - - B
,i[l:;;;;:::::J~::::::;;;:::;;=:;:::;=~O~.l~mt
0.184 mm
SF
0.286 mm
{d)
Fig. 9-7
'*
( -o
-102 10 ) m
Ans.
Nso = 30kN
N11c = (i() kN
(c)
420
I
CHAPTER
EXAMPLE
9
AXIAL L OAD
9.4
A member is made of a material that has a specific weight of y = 6 kN / m3
and modulus of elasticity of 9 GPa. If it is in the form of a cone having the
dimensions shown in Fig. 9-&1, determine how far its end is displaced due to
gravity when it is suspended in the vertical position.
Y
SOLUTION
The internal axial force varies along the member, since
it is dependent on the weight W(y) of a segment of the member below
any section, Fig. 9-8b. Hence, to calculate the displacement, we must use
Eq. 9- 1. At the section located a distance y from the cone's free end, the
radius x of the cone as a function of y is determined by proportion; i.e.,
Internal Force.
3m
x
-
y
=
0.3m
3m '
x = O.ly
The volume of a cone having a base of radius x and height y is
V
=
1
31Tyx 2
=
(a)
Since W
=
?T(0.01) 3
y
3
I
N(y)
=
N(y)
O·,
=
6(103 )(0.01047y3 )
=
62.83y3
The area of the cross section is also a function of
position y, Fig. 9-8b. We have
Displacement.
A(y)
=
2
?TX
0.03142y2
=
Applying Eq. 9- 1 between the limits of y
(b)
0.01047y3
yV, the internal force at the section becomes
y
+ jIF.y
=
o
=
(LN(y)dy
} 0 A(y) E
Fig. 9-8
0 and y
=
=
3 m yields
1
1
3
=
(62.83y3 ) dy
o (0.03142y2 ) 9(109)
3
9
=
222.2(10-
=
1(10- 6 ) m
)
=
y dy
1 µm
NOTE: This is indeed a very small amount.
Ans.
9.2
421
ELASTIC DEFORMATION OF AN AxlALLY LOADED M EMBER
PRELIMINARY PROBLEMS
P9-l . In each case. determine the internal normal force
between lettered points on the bar. Draw all necessary freebody diagrams.
I'
8
c
D
P9-4. The rod is subjected to an external axial force of
800 Nanda uniform distributed load of 100 N/m along its
length. Determine the internal normal force in the rod as a
function of x.
'
(a)
A
C
8
D
600N --+:f\f>~;;;;;;;;~4'00~N~~~~~300~~~~.~~~~.
(b)
Prob. P9-l
IOON/m
A
~~~~i,5~~Sd~~ 800N
1- - -- - - 2 m
-
1---x-i
I
Prob. P9-4
P9-2. Determine the internal normal force between
lettered points on the cable and rod. Draw all necessary
free-body diagrams.
P9-5. The rigid beam supports the load of 60 kN.
Determine the displacement at 8. Take E = 60 GPa. and
Asc = 2 (10-3) m 2.
SOON
•
8
D
Prob. P9-2
P9-3. The post weighs 8 kN/m. Determine the internal
normal force in the post as a function of x.
60kN
12m--4m-1
A
TI
8
T
T
2 111
3m
Jo
2m
Prob. P9-3
c
Prob.P9-5
422
CHAPTER
9
AXIAL L OAD
FUNDAMENTAL PROBLEMS
F9-1. The 20-mm-diameter A-36 steel rod is subjected to
the axial forces shown. Determine the displacement of
end C with respect to the fixed support at A.
-
600 mm, -400 mm - J
~
~
A
!1--4-;0.~kN
SOkN
nU SOkN c
F9-4. If the 20-mrn-diameter rod is made of A-36 steel
and the stiffness of the spring is k = 50 MN/m, determine
the displacement of end A when the 60-kN force is applied.
400mm
k = SOMN/m
I
Prob.F9-l
400mm
J_
F9-2. Segments AB and CD of the assembly are solid
circular rods, and segment BC is a tube. If the assembly is
made of 6061-T6 aluminum, determine the displacement of
end D with respect to end A.
A
60kN
Prob.F9-4
20mm
-
lO kN
Al
lOkN-
a
B
c
lSkN D
: Ill 20kN
I
1
1
tokN
400mm
a
400mm
lSkN I
400mm
I
F9-5. The 20-mrn-diameter 2014-T6 aluminum rod is
subjected to the uniform distributed axial load. Determine
the displacement of end A.
t~kN/m
A l~~~~~~~
~
30mmIO 140mm
1 - - - - - -900 mm _ _ _ _ __,
Section a-a
Prob.F9-2
Prob. F9-5
F9-3. The 30-mrn-diameter A992 steel rod is subjected to
the loading shown. Determine the displacement of end C.
F9-6. The 20-mrn-diameter 2014-T6 aluminum rod is
subjected to the triangular distributed axial load. Determine
the displacement of end A.
45 kN/m
s /
30kN
~3
A
I
B ~3
90 kN
c
I ~ """-30 kN
1 -400 mm - - i - - -600 mm- - - 1
Prob.F9-3
1 - - - - - -900 mm - - - - --
Prob.F9-6
9.2
423
ELASTIC DEFORMATION OF AN AxlALLY LOADED MEMBER
PROBLEMS
9-L The A992 steel rod is subjected to the loading shown.
If the cross-sectional area of the rod is 80 mm2, determine the
displacement of B and A. Neglect the size of the couplings
at Band C.
9- 3. The composite shaft, consisting of aluminum, copper,
and steel sections, is subjected to the loading shown.
Determine the displacement of end A with respect to end D
and the normal stress in each section. The cross-sectional
area and modulus of elasticity for each section are shown in
the figure. Neglect the size of the collars at B and C.
*9-4. The composite shaft, consisting of aluminum, copper,
and steel sections, is subjected to the loading shown. Determine
the displacement of B with respect to C. The cross-sectional
area and modulus of elasticity for each section are shown in
the figUie. Neglect the size of the collars at Band C.
I
0.75m
l
c
1.50 m
6kN
6 kN
Aluminum
£ 31 = 10(Hl3) ksi
B
AA 8
I
lm
5 kN
2.~
l
SkN
A
= 0.09 in2
A
Aco = 0.06 in2
n : fr3.50 kip
1.75 kip
1; 0 kip
·~.50 c 11"
kip
f
kip
16 in.- !
- -18 in.- l -12 in.lOkN
Steel
E,. = 29(103 ) ksi
Copper
£ 00 = 18(103 ) ksi
A8 c = 0.12 in2
Probs. 9- 314
Prob. 9- 1
9-2. The copper shaft is subjected to the axial loads shown.
Determine the displacement of end A with respect to end D
if the diameters of each segment are dA 8 = 0.75 in., d 8 c =
1 in., and dcD = 0.5 in. Take Ecu = 18(HP) ksi.
A 8 kN
- 80 in.- j -150 in.- j -100 in.-J
s kip
8 kip
..A
I
I
I 2 kip
It
:
5 kip B
C 2kip
Prob. 9- 2
I
D
9-5. The 2014-T6 aluminum rod has a diameter of 30 mm
and supports the load shown. Determine the displacement of
end A with respect to end £.Neglect the size of the couplings.
..
6 kip
B
C
D
E
o..fktD ttD 0~kN I
1-4m~-2m-l-2m-1-2mProb. 9- 5
424
C HAPT ER
9
A XIAL L OAD
9-6. The A-36 steel drill shaft of an oil well extends 12 000 ft
into the ground. Assuming that the pipe used to drill the well
is suspended freely from the derrick at A , determine the
maximum average normal stress in each pipe string and the
elongation of its end D with respect to the fixed end
at A. The shaft consists of three different sizes of pipe, AB,
BC, and CD, each having the length, weight per unit length,
and cross-sectional area indicated.
9-9. The assembly consists of two 10-mm diameter red
brass C83400 copper rods AB and CD , a 15-mm diameter
304 stainless steel rod EF, and a rigid bar G. If P = 5 kN,
determine the horizontal displacement of end F of rod EF.
9-10. The assembly consists of two 10-mm diameter red
brass C83400 copper rods AB and CD , a 15-mm diameter
304 stainless steel rod EF, and a rigid bar G. If the horizontal
displacement of end F of rod EF is 0.45 mm, determine the
magnitude of P.
450mm -
i - 300mm AAB =
2.50 in.2
IYAB =
3.2 lb/ft
B-
A
-
p
...
E
•
A8 c = 1.75 in.2
w8 c = 2.8 lb /ft
c
D ~
4P •
F-
p
G
Probs. 9-9/10
1.25 in.2
IYCD = 2.Q lb /ft
Arn =
Prob. 9-6
9-7. The truss is made of three A-36 steel members, each
having a cross-sectional area of 400 mm2 . Determine the
horizontal displacement of the roller at C when P = 8 kN.
*9-8. The truss is made of three A-36 steel members,
each having a cross-sectional area of 400 mm2 . Determine
the magnitude P required to displace the roller to the
right 0.2 mm.
9-11. The load is supported by the four 304 stainless steel
wires that are connected to the rigid members AB and DC.
Determine the vertical displacement of the 500-lb load if
the members were originally horizontal when the load was
applied. Each wire has a cross-sectional area of 0.025 in2 .
*9-12. The load is supported by the four 304 stainless steel
wires that are connected to the rigid members AB and DC.
Determine t he angle of tilt of each member after the 500-lb
load is applied. The members were originally horizontal,
and each wire has a cross-sectional area of 0.025 in2 •
E
---·
'"
-·
F
- .-
..
G
~
-
-
p
3 ft
D
T
0.8m
5 ft
H
c
1- 1 ft- -
2ft - I
1.8 ft
A~/<-----------~
o~--i------=hJ,_
1 - - -0.8
m~--0.6 m- 1
Probs. 9-7/8
I
I
A
B
I
3 ft
- 1 ft-I
500 lb
Probs. 9-11112
9.2
ELASTIC DEFORMATION OF AN AxlALLY LOADED M EMBER
9-13. The rigid bar is supported by the pin-connected rod
CB that has a cross-sectional area of 14 mm2 and is made
from 6061-T6 aluminum. Determine the vertical deflection
of the bar at D when the distributed load is applied.
4 25
*9-16. The coupling rod is subjected to a force of 5 kip.
Determine the distance d between C and E accounting for
the compression of the spring and the deformation of the
bolts. When no load is applied the spring is unstretched and
d = 10 in. The material is A-36 steel and each bolt has a
diameter of 0.25 in. The plates al A. B, and Care rigid and the
spring has a stiffness or k = 12 kip/ in.
-1
1.5 m
L~~____.D
Prob. 9-13
9-14. The post is made of Douglas fir and has a diameter
of 100 mm. If it is subjected to the load of20 kN and the soil
provides a frictional resistance distributed around the post
that is triangular along its sides; that is, it varies from w = 0
at y = 0 tow = 12 kN/m at y = 2 m, determine the force
F at its bottom needed for equilibrium. Also, what is the
displacement of the top of the post A with respect to its
bottom B? Neglect the weight of the post.
9-15. The post is made of Douglas fir and has a diameter
of 100 mm. if it is subjected to the load of20 kN and the soil
provides a frictional resistance that is distributed along
its length and varies linearly from IV= 4 kN/ m at y = 0
to IV =12 kN/ m at y = 2 m. determine the force F at its
bottom needed for equilibrium. Also, what is the
displacement or the top of the post A with respect to its
bottom B? Neglect the weight of the post.
i
S kip
Prob.9-16
9-17. The pipe is stuck in lhe ground so that when it is
pulled upward the frictional force along its length varies
linearly from zero al B to /mu (force/length) at C. Determine
the initial force P required to pull the pipe out and the
pipe's elongation just before it starts to slip. The pipe has a
length L. cross-sectional area A, and the material from
which it is made has a modulus of elasticity £.
p
20kN
y
I
2mj_
+ +
t
t
t t
t t
IV
12 kN/m
!""'.
Probs. 9-14115
c
Prob. 9-17
426
9
CHAPTER
AXIAL L OAD
9-18. The linkage is made of three pin-connected A992
steel members, each having a diameter of 1~ in. If a
horizontal force of P = 60 kip is applied to the end B of
member AB, determine the displacement of point B.
9-22. The rigid beam is supported at its ends by two A-36
steel tie rods. The rods have diameters dAa = 0.5 in. and
dcv = 0.3 in. If the allowable stress for the steel is
u allow = 16.2 ksi, determine the largest intensity of the
distributed load w and its length x on the beam so that the
beam remains in the horizontal position when it is loaded.
-I
4 ft
p
l
-I
I---6
4 ft
ft
9-21. The rigid beam is supported at its ends by two A-36
steel tie rods. If the allowable stress for the steel is
uauaw= 16.2 ksi, the load w=3 kip/ft, andx=4 ft, determine
the smallest diameter of each rod so that the beam remains
in the horizontal position when it is loaded.
---ii B
_l
Prob.9-18
B
.J
•
9-19. The linkage is made of three pin-connected A992
steel members, each having a diameter of 1~ in. Determine
the magnitude of the force P needed to displace point B
0.25 in. to the right.
I :i
A..,.
~8ft
4 ft
l
-I
p
- - 6 ft - - -118
4 ft
1
l
6ft
JV
-I
D
I
c
I
Probs. 9-21122
_l
Prob.9-19
*9-20. The assembly consists of three titanium (Ti-6Al-4V)
rods and a rigid bar AC. The cross-sectional area of each rod
is given in the figure. If a force of 60 kip is applied to the
ring F, determine the horizontal displacement of point F.
A
A £F = 2 .lD2
60k ip
6 ft
T
1 ft
9-23. The steel bar has the original dimensions shown in
the figure. If it is subjected to an axial load of 50 kN,
determine the change in its length and its new cross-sectional
dimensions at section a-a. E" = 200 GPa, v" = 0.29.
B,
A A8 = 1 in2
'
A co = 1.5 in2
'
E
i l -2ft
2 ft
-~
1
c
Prob.9-20
6 ft
SOkN
D"
Prob.9-23
9.2
ELASTIC DEFORMATION OF AN AxlALLY LOADED M EMBER
*9-24. Determine the relative displacement of one end of
the tapered plate with respect to the other end when it is
subjected to an axial load P.
p
/ - -d2----I
427
*9-28. The observation cage Chas a weight of 250 kip and
through a system of gears, travels upward at constant
velocity along the A-36 steel column, which has a height of
200 ft. The column has an outer diameter of 3 ft and is made
from steel plate having a thickness of 0.25 in. Neglect the
weight of the column, and determine the average normal
stress in the column at its base. 8. as a function of the cage's
position y. Also, determine the displacement of end A as a
function of y.
A
200 ft
Prob.9-24
9-25. The assembly consists of two rigid bars that are
originally horizontal. They arc supported by pins and
0.25-in.-diamctcr A-36 steel rods. If the vertical load of 5 kip
is applied to the bottom bar A 8, determine the displacement
at C, 8 , and£.
c
1
~aooooaliQ
y
...L_
B
.
•f
~
Ji
J.
1-2 rt -I,.
8 ft
c
A
•
£/
•
B
I- ri---1f1•'>
6
-6
r1 -I
bl.5 rtD
i
Prob. 9-28
9-29. Determine the elongation of the aluminum strap
when it is subjected to an axial force of 30 kN. E.1 =70 GPa.
30 lcN
5 kip
Prob.9-25
9-26. The truss consists of three members, each made
from A-36 steel and having a cross-sectional area of 0.75 in2 •
Detemune the greatest load P that can be applied so that
the roller support at 8 is not displaced more than 0.03 in.
9-27. Solve Prob. 9-26 when the load P acts vertically
downward at C.
c
--·~---800 mm ---il---11
~250mm
250mm
Prob. 9-29
9-30. The ball is truncated at its ends and is used to
support the bearing load P. If the modulus of elasticity for
the material is£, determine the decrease in the ball's height
when the load is applied.
p
r
2
r
Probs. 9-26127
30kN
Prob. 9-30
428
CHAPTER
9
AXIAL L OAD
9.3
PRINCIPLE OF SUPERPOSITION
The principle of superposition is often used to determine the stress or
displacement at a point in a member when the member is subjected to a
complicated loading. By subdividing the loading into components, this
principle states that the resultant stress or displacement at the point can
be determined by algebraically summing the stress or displacement
caused by each load component applied separately to the member.
The following two conditions must be satisfied if the principle of
superposition is to be applied.
1. The loading N must be linearly related to the stress u or
displacement B that is to be determined. For example, the equations
a = N /A and 5 = NL/ AE involve a linear relationship between
a and N, and 5 and N.
2. The loading must not significantly change the original geometry or
configuration of the member. If significant changes do occur, the
direction and location of the applied forces and their moment arms
will change. For example, consider the slender rod shown in Fig. 9- 9a,
which is subjected to the load P. In Fig. 9- 9b, P is replaced by two
of its components, P = P 1 + P2 . If P causes the rod to deflect a large
amount, as shown, the moment of this load about its support, Pd, will
not equal the sum of the moments of its component loads,
Pd # P1d 1 + P2d 2, because d 1 # d 2 # d.
*
~
P,
d,
J
P2
+
di
d
(a)
(b)
Fi.g. 9-9
9.4
STATICALLY INDETERMINATE
AXIALLY LOADED MEMBERS
Consider the bar shown in Fig. 9- lOa, which is fixed supported at both of
its ends. From its free-body diagram, Fig. 9- lOb, there are two unknown
support reactions. Equilibrium requires
+ f2F
=
O;
FB + FA - 500 N
=
0
This type of problem is called statically indeterminate, since the equilibrium
equation is not sufficient to determine both reactions on the bar.
9.4
In order to establish an additional equation needed for solution, it is necessary
to consider how points on the bar are displaced. Specifically, an equation that
specifies the conditions for displacement is referred to as a compatibility or
kinematic condition. In this case, a suitable compatibility condition would require
the displacement of end A of the bar with respect to end B to equal zero, since the
end supports are fixed , and so no relative movement can occur between them.
Hence, the compatibility condition becomes
{,A ja
4 29
STATICALLY INDETERMINATE AxlALLY LOADED MEMBERS
A
I2 m
t-
= Q
c
SOON
3m
This equation can be expressed in terms of the internal loads by using a
loa~isplacement relationship, which depends on the material behavior. For
example, if linear elastic behavior occurs, then l'J = NL/AE can be used. Realizing
that the internal force in segment AC is +FA , and in segment CB it is -Fa, Fig. 9-lOc,
then the compatibility equation can be written as
FA(2 m)
Fa(3 m)
AE AE
=
(a)
O
Since AE is constant, t he n FA= 1.5F8 . Finally, using t he equilibrium equation, the
reactions are therefore
FA
= 300 N
and
Fa
= 200 N
SOON
Since both of these results are positive, the directions of the reactions are shown
correctly on the free-body diagram.
To solve for the reactions on any statically indeterminate problem, we must
therefore satisfy both the equilibrium and compatibility equations, and relate the
displacements to the loads using the load-displacement relations.
IMPORTANT POINTS
• The principle of superposition is sometimes used to simplify
stress and displacement problems having complicated loadings.
This is done by subdividing the loading into components, then
algebraically adding the results.
• Superposition requires that the loading be linearly related to the
stress or displacement, and the loading must not significantly
change the original geometry of the member.
• A problem is statically indeterminate if the equations of
equilibrium are not sufficient to determine all the reactions on
a member.
• Compatibility conditions specify the displacement constraints
that occur at the supports or other points on a member.
i
(b)
Fig. 9-10
(c)
430
CHAPTER
9
AXIAL L OAD
PROCEDURE FOR ANALYSIS
The support reactions for statically indeterminate problems
are determined by satisfying equilibrium, compatibility, and
load- displacement requirements for the member.
Equilibrium.
• Draw a free-body diagram of the member in order to identify
all the forces that act on it.
• The problem can be classified as statically indeterminate if the
number of unknown reactions on the free-body diagram is
greater than the number of available equations of equilibrium.
• Write the equations of equilibrium for the member.
Most concrete columns are reinforced with
steel rods; and since these two materials work
together in supporting the applied load, the
force in each material must be determined
using a statically indeterminate analysis.
Compatibility.
• Consider drawing a displacement diagram in order to
investigate the way the member will elongate or contract when
subjected to the external loads.
• Express the compatibility conditions in terms of the
displacements caused by the loading.
Load-Displacement.
• Use a load- displacement relation, such as a = NL/ AE, to
relate the unknown displacements in the compatibility
equation to the reactions.
• Solve all the equations for the reactions. If any of the results
has a negative numerical value, it indicates that this force
acts in the opposite sense of direction to that indicated on the
free-body diagram.
9.4
I
EXAMPLE
STATICALLY INDETERMINATE AxlALLY LOADED MEMBERS
431
9.5
The steel rod shown in Fig. 9- lla has a diameter of 10 mm. It i.s fixed
to the wall at A , and before it is loaded, there is a gap of 0.2 mm
between the wall at B' and the rod. Determine the reactions on the
rod if it is subjected to an axial force of P = 20 kN. Neglect the size of
the collar at C. Take £,1 = 200 GPa.
p = 20 kN
A
0.2 mm~
$ Ea:O.::=:=:E[j'I B'
1.:_800 mm ___!J
400 mm
(a)
SOLUTION
Equilibrium. As shown on the free-body diagram, Fig. 9- 1 lb, we
will assume that force P is large enough to cause the rod's end B to
contact the wall at B'. When this occurs, the problem becomes
statically indeterminate since there are two unknowns and only one
equation of equilibrium.
~ 'I.fr = 0;
-FA - FB + 20(103) N
=
0
(1)
Compatibility. The force P causes point B to move to B; with no
further displacement. Therefore the compatibility condition for the rod is
oB/A
FA ~......-----;~FA
0.0002 m
=
Load-Displacement. This displacement can be expressed in terms
of the unknown reactions using the load-displacement relationship,
Eq. 9-2, applied to segments AC and CB, Fig. 9- llc. Working in units
of newtons and meters, we have
FA LAc
OB/A =
FBLcB
AE
AE
=
0.0002m
FA(0.4 m)
7T(0.005 m) 2 (200(109) N/m2]
FB (0.8 m)
----------- =
7T(0.005 m) 2 (200(10 9) N/m 2]
0.0002 m
or
FA (0.4 m) - FB (0.8 m)
=
3141.59 N • m
(2)
Solving Eqs. 1 and 2 yields
FA = 16.0kN
FB
=
4.05 kN
Ans.
Since the answer for FB is positive, indeed end B contacts the wall at
B' as originally assumed.
NOTE: If FB were a negative quantity, the problem would be
statically determinate, so that FB
=
0 and FA = 20 kN.
Fs -~;;;;;;;;~~g..- F8
(c)
Fig. 9-11
432
I
CHAPTER
EXAMPLE
9
AXIAL L OAD
9.6
The aluminum post shown in JFig. 9- l2a is reinforced with a brass core.
If this assembly supports an axial compressive load of P = 9 kip,
applied to the rigid cap, determine the average normal stress in the
aluminum and the brass. Take Ea1 = 10(103) ksi and Ebr = 15(103) ksi.
2 in.
1.5 ft
SOLUTION
Equilibrium. The free-body diagram of the post is shown in Fig. 9- 12b.
Here the resultant axial force at the base is represented by the unknown
components carried by the aluminum, Fah and brass, Fbr· The problem is
statically indeterminate.
Vertical force equilibrium requires
(a)
+ jIFy
!P=
9kip
-9kip + Fal + Fbr
O·,
=
(1)
0
Compatibility. The rigid cap at the top of the post causes both the
aluminum and brass to be displaced the same amount. Therefore,
Bal
F.1 L
F
=
al
[
Fbr
=
Bbr
Using the load-displacement relationships,
Load-Displacement.
(b)
=
FbrL
2
7T((2in.) - (lin.)
7T(l in.) 2
F.1
=
2
]
J
3
[10(10 )ksi ]
15(103) ksi
(2)
2fbr
Solving Eqs. 1 and 2 simultaneously yields
Fa1
6 kip
=
Jbr
=
3 kip
The average normal stress in the aluminum and brass is therefore
6 kip
CTal =
.
)2
7T [(2 m.
3 kip
Obr =
(c)
Fig. 9-U
7T(l in.) 2
=
.
(
.
1 m.)
2] =
0.955 ksi
0.637
ks1
Ans.
Ans.
NOTE: Using these results, the stress distributions within the materials
are shown in Fig. 9- 12c. Here the stiffer material, brass, is subjected to
the larger stress.
9.4
EXAMPLE
433
STATICALLY INDETERMINATE AxlALLY LOADED M EMBERS
9.7
The three A992 steel bars shown in Fig. 9- 13a are pin connected to a
rigid member. lf the applied load on the member is 15 kN, determine
the force developed in each bar. Bars AB and EF each have a
cross-sectional area of 50 mm2, and bar CD has a cross-sectional area
of30 mnl.
SOLUTION
•
•
~
~
••
• 1
B.
F;
D"
05m
c.
A.
l
E l.
•
Equilibrium. The free-body diagram of the rigid member is shown
in Fig. 9-13b. This problem is statically indeterminate since there are
three unknowns and only two available equilibrium equations.
I
0.4m-
0.2m l0.2m
I ,)
FA + Fc + FE - 15 kN
+ jIF,, = O;
~+"I.Mc
= O;
=
(1)
0
-FA(0.4 m) + 15 kN(0.2 m) + F E(0.4 m)
=0
(2)
15 kN
(a)
Compatibility. The applied load will cause the horizontal line ACE
shown in Fig. 9-13c to move to the inclined position A' C' E'. The red
displacements C>A, Cle, C>E can be related by similar triangles. Thus the
compatibility equation that relates these displacements is
FA
t
Fe
F£
_]t~
g l0.2ml 0.4m
15 kN
Load- Displacement.
Eq. 9-2, we have
Using the load-displacement relationship,
(b)
Ar-0.4
FA L
I [
(30
mm2)£
51
=2
(50
mm2)£ SI
]
1 [
+2
Fe = 0.3FA + 0.3FE
8E~
FE L
(50 rrun2)£ SI ]
md- md
0.4
8,. -8E
(3)
A' ,. 8c - 8E C' 8c
8
(c)
Solving Eqs.1-3 simultaneously yields
Fig. 9-13
FA= 9.52 kN
Ans.
Fe= 3.46 kN
Ans.
FE= 2.02kN
Ans.
l 8E
E'
434
I
CHAPTER
EXAMPLE
9
AXIAL L OAD
9.8
The bolt shown in Fig. 9- l4a is made of 2014-T6 aluminum alloy, and it
passes through the cylindrical tube made of Am 1004-T61 magnesium
alloy. The tube has an outer radius of! in., and it is assumed that both the
inner radius of the tube and the radius of the bolt are l in. When the bolt
is snug against the tube it produces negligible force in the tube. Using a
wrench, the nut is then further tightened one-half turn. If the bolt has
20 threads per inch, determine the stress in the bolt.
I .
2m.
SOLUTION
Equilibrium. The free-body diagram of a section of the bolt and the
tube, Fig. 9- 14b, is considered in order to relate the force in the bolt Fb to
that in the tube, F,. Equilibrium requires
(a)
+ jIF.y
=
F,,-F, = 0
O·,
(1)
F,
Compatibility. As noted in Fig. 9- 14c, when the nut is tightened onehalf turn on the bolt, it advances a distance of (~) {z10 in.) = 0.025 in.
This will cause the tube to shorten 5 1 and the bolt to elongate 5 b· Thus,
the compatibility of these displacements requires
11.
'
(+j)
5,
+ 5b
=
0.025 in.
Load-Displacement. Taking the moduli of elasticity from the table on
the inside back cover, and applying the load-displacement relationship,
Eq. 9- 2, yields
F, (3 in.)
''l
ff!
(b}
7T[(0.5 in.)2
-
(0.25 in.)2) (6.48(103) ksi)
+
Fb (3 in.)
11(0.25 in.)2 (10.6(1G3) ksi)
0.78595F,
+
=
0.025 in.
(2)
1.4414Fb = 25
Solving Eqs. 1 and 2 simultaneously, we get
Fb
Final
position
0.025 in.
Initial
position
(c)
Fig. 9-14
=
F,
11.22 kip
=
The stresses in the bolt and tube are therefore
Fb
11.22 kip
<Tb = =
= 57.2 ksi
Ab
7T(0.25 in.) 2
F,
a: = '
A,
-
11.22 kip
7T[(0.5 in.)2
-
(0.25 in.)2 )
Ans.
=
19.1 ksi
These stresses are Jess than the reported yield stress for each material,
(uv)a1 = 60 ksi and (uv)mg = 22 ksi (see the inside back cover), and
therefore this "elastic" analysis is valid.
9.5
9. 5
4 35
THE FORCE METHOD OF ANALYSIS FOR AxlALLY LOADED M EMBERS
THE FORCE METHOD OF ANALYSIS
FOR AXIALLY LOADED MEMBERS
It is also possible to solve statically indeterminate problems by writing the
compatibility equation using the principle of superposition. This method
of solution is often referred to as the flexibility or force method of
analysis. To show how it is applied, consider again the bar in Fig. 9-15a.
U we choose the support at B as "redundant" and temporarily remove it
from the bar, then the bar will become statically determinate, as in
Fig. 9-15b. Using the principle of superposition, however, we must add
back the unknown redundant load F8 , as shown in Fig. 9-15c.
Since the load P causes B to be displaced downward by an amount 8p,
the reaction F8 must displace end B of the bar upward by an amount 88 ,
so that no displacement occurs at B when the two loadings are
superimposed. Assuming displacements are positive downward, we have
A
I2m
No dasploc:ement at B
c
SOON
(•)
I
Jm
1
B
II
A
Oi.splaccme.n1 Ill 8 when
redw1dan1 force at 8
is rcn10\ Cd
1
(b)
500 N
(+!)
This condition of 8p = 8 8 represents the compatibility equation for
displacements at point B.
Applying the load-displacement relationship to each bar, we have
8p = 500 N(2 m)/AE and 88 = F8 (5 m) /AE. Consequently,
O=
500 N ( 2 m )
Fs ( 5 m)
AE
AE
+
A
Oisplacen\Cnt 111 8 when
onl)' the redundanl force
l
ot Bis opplied
Fs = 200N
(c)
500N
From the free-body diagram of the bar, Fig. 9- 15d, equilibrium requires
+f !F1 = O;
200N +FA - SOON = 0
Fa
Then
(d)
FA= 300N
Fig. 9-15
These results are the same as those obtained in Sec. 9.4.
PROCEDURE FOR ANALYSIS
The force method of analysis requires the following steps.
Compatibility.
• Choose one of the supports as redundant and write the equation of compatibility. To do this, the known
displacement at the redundant support, which is usually zero, is equated to the displacement at the support
caused only by the external loads acting on the member plus (vectorially) the displacement at this support
caused only by the redundant reaction acting on the member.
436
CHAPTER
9
AXIAL L OAD
Load-Displacement.
• Express the external load and redundant displacements in terms of t he loadings by usmg a
load-displacement relationship, such as a = NL/ AE.
• Once established, the compatibility equation can then be solved for the magnitude of the redundant force.
Equilibrium.
• D raw a free-body diagram and write the appropriate equations of equilibrium for the member using the
calculated result for the redundant. Solve these equations for the other reactions.
I
EXAMPLE
9.9
TheA-36 steel rod shown in Fig. 9-16a has a diameter of 10 mm. It is fixed to
the wall at A , and before it is loaded there is a gap between the wall and the
rod of 0.2 mm. Determine the reactions at A and B '. Neglect the size of the
collar at C. Take E 51 = 200 GPa.
0.2nm~
B'
800mm -
P = 20kN
P = 20 kN
SOLUTION
11
- ( •3
Compatibility. Here we will consider the support at B' as redundant.
Using the principle of superposit ion, Fig. 9- 16b, we have
Initial~ I
position 1-c.Sp
- i- l
f NAc =20kN Ncs=O
+
kfinal
~/ls position
~=:3;~;;;;;;;=:=;;-i;i-t-(b)
(~)
Fs
0.0002 m =
op -
(1)
DB
Load-Displacement. The defiections Op and SB are determined from
Eq. 9- 2.
NAcLAc
op =
AE
-
(20(10 3 ) N)(0.4 m)
_3
5
93
?T(0.005 m) 2 (200(109) N/m2) = 0. 0 (lO ) m
-
FB (1.20 m)
---~--~-~
= 76.3944(10- 9)FB
2
9
2
?T(0.005 m) (200(10
)
N/m
]
Substituting into Eq. 1, we get
0.0002 m = 0.5093(10- 3 ) m - 76.3944(10- 9 )FB
FB
(c)
Fig. 9- 16
=
4.05(103 ) N = 4.05 kN
Ans.
Equilibrium. From the free-body diagram, Fig. 9- 16c,
+ "'F.x
~ ~
=
O·'
-FA+ 20kN - 4.05kN
=
0
FA = 16.0kN
Ans.
9.5
437
THE FORCE M ETHOD OF ANALYSIS FOR AxlALLY LOADED M EMBERS
PROBLEMS
9-31. The column is constructed from high-strength
concrete and eight A992 steel reinforcing rods.Uthe column
is subjected to an axial force of 200 kip, determine the
average normal stress in the concrete and in each rod. Each
rod has a diameter of l in.
*9-32. The column is constructed from high-strength
concrete and eight A992 steel reinforcing rods.Uthe column
is subjected to an axial force of 200 kip, determine the
required diameter of each rod so that 60% of the axial force
is carried by the concrete.
9-34. If column AB is made Crom high strength precast
concrete and reinforced with four : in. diameter A-36 ste.el
rods, determine the average normal stress developed in the
concrete and in each rod. Set P = 75 kip.
9-35. If column AB is made Crom high strength precast
concrete and reinforced with four in. diameter A-36 steel
rods, determine the maximum allowable floor loadings P.
The allowable normal stresses for the concrete and the steel
are (uanow)con = 2.5 ksi and (ua11ow)s1 =24 ksi, respectively.
l
p
4
p
0
Ill.
•
•
200 kip
I: :J=r9 in.
a -- --a
~I
9 in.
1111
1111
I Ii I
JIJ I
1111
IJIJ
I JI I
IJI I
1111 IIIi
1111 lili
1111 lili
I Ii I I JI I
Jiii 1111
1111 IJI I
Jiii llli
Jiii 1111
JI JI IJI I
I Ii I I JI I
to ft
Section a-a
3 ft
8
Probs. 9-34135
*9-36. Determine the support reactions at the rigid supports
A and C. The material has a modulus of elasticity of£.
Probs. 9-31/32
9-33. The A-36 steel pipe has a 6061-T6 aluminum core. It
is subjected to a tensile force of 200 kN. Determine the
average normal stress in th e aluminum and the steel due to
this loading. The pipe has an outer diameter of 80 mm and
an inner diameter of 70 mm.
9-37. If the supports at A and C are flexible and have a
stiffness k, determine the support reactions at A and C. The
material has a modulus of elasticity of£.
,.
1-,
.l.t1
ti
I
41
p
I
200kN <I
400mm
02
I
)
Prob. 9-33
A
200kN
I
;
I
8
2a
Probs. 9-36/37
I
a
c
438
9
CHAPTER
AXIAL L OAD
9-38. The load of 2000 lb is to be supported by the two
vertical steel wires for which uy = 70 ksi. Originally wire AB
is 60 in. long and wire AC is 60.04 in. long. Determine the force
developed in each wire after the load is suspended. Each wire
has a cross-sectional area of0.02 in2 . E" = 29.0(lQ'.l) ksi.
9-39. The load of 2000 lb is to be supported by the two
vertical steel wires for which uy = 70 ksi. Originally wire
AB is 60 in. Jong and wire AC is 60.04 in. long. Determine
the cross-sectional area of AB if the load is to be shared
equally between both wires. Wire AC has a cross-sectional
area of 0.02 in2 . Esi = 29.0(103) ksi.
9-4L The 10-mm-diameter steel bolt is surrounded by a
bronze sleeve. The outer diameter of this sleeve is 20 mm, and
its inner diameter is 10 mm. If the yield stress for the steel is
(uy)si = 640 MPa, and for the bronze (uy)br = 520 MPa,
determine the magnitude of the largest elastic load P that can
be applied to the assembly.£"= 200 GPa, Eb,= 100 GPa.
p
I
I
-
-
I
I
I
I
I
I
lOmm
1
I
I
B
c
--20mm
I
•• '. '
ii•
t
60.04 in.
p
Prob. 9-41
9-42. The 10-mm-diameter steel bolt is surrounded by a
bronze sleeve. The outer diameter of this sleeve is 20 mm, and
its inner diameter is 10 mm. If the bolt is subjected to a
compressive force of P=20 kN,determine the average normal
stress in the steel and the bronze.£.,= 200 GPa, Ebr = 100 GPa.
Probs. 9-38/39
p
*9-40. The A-36 steel pipe has an outer radius of 20 mm
and an inner radius of 15 mm. If it fits snugly between the
fixed walls before it is loaded, determine the reaction at the
walls when it is subjected to the load shown.
I
I
I
- - +'I
I
I
I
lOmm
- -1
- -20mm
I
I
A
B
l-300mm-I
-
8kN
--
8kN
700mm
Prob. 9-40
c
'
I
I
p
Prob. 9-42
9.5
THE FORCE METHOD OF ANALYSIS FOR AxlALLY LOADED M EMBERS
9-43. The assembly consists of two red brass C83400
copper rods AB and CD of diameter 30 mm, a stainless 304
steel alloy rod EF of diameter 40 mm, and a rigid cap G. If
the supports at A, C. and Fare rigid, determine the average
normal stress developed in the rods.
If the gap between C and the rigid wall at D is
initially 0.15 mm, determine the support reactions at A and
D when the force P = 200 kN is applied. The assembly is
made of solid A-36 steel cylinders.
9-46.
1600 --i-600
mm
I
•
I
A
I
30mm
E
I
4
c 30mm
r
D •
G
A
F
I
p
40kN
1._)
*9-44. The rigid beam is supported by the three suspender
bars. Bars AB and EF arc made of aluminum and bar CD is
made of steel. lf each bar has a cross-sectional area of 450 mm2 ,
determine the maximum va lue of P if the allowable stress is
(uauow)si = 200 MPa for the steel and (<rauow)a1 = 150 MPa for
the aluminum.£,,= 200 GPa, Ea1=70 GPa.
'•
0
1>
~
2m
al
st
,_ O.ISmm
I
I
25mm
B
I
D
c
,.
Prob. 9-46
9-47. The support consists of a solid red brass C83400
copper post surrou11ded by a 304 stainless steel tube. Before
the load is applied the gap between these two parts is 1 mm.
Given the dimensions s hown, determine the greatest axial
load that ca11 be applied to the rigid cap A without causing
yielding of any one of the materials.
p
.
r~~
. +lmm
F
D
B
SO.Imm
I
40mm
Prob.9-43
al
mml
4 40kN
B
439
0.25 m
•
'
'
c
A
I
0
E
.n
0
0.75m 0.15 m '0.75 m 0.75m
p
~>-IOmm
mm
au
'
2P
Prob. 9-44
9-45. The bolt AB has a diameter of 20 mm and passes
through a sleeve that has an inner diameter of 40 mm and
an outer diameter of 50 mm. The bolt and sleeve are made
of A-36 steel and are secured to the rigid brackets as shown.
If the bolt length is 220 mm and the sleeve length is 200 mm,
determine the tension in the bolt when a force of 50 kN is
applied to the brackets.
l-200mm-I
::: :A
9il-220mm-J
- [el::::
Prob. 9-45
Prob.9-47
*9-48. The specimen represents a filament-reinforced matrix
system made from plastic (matrix) and glass (fiber). If there are
11 fibers.. each having a cross-sectional area of A and modulus
1
of Er embedded in a matrix having a cross-sectional area of Am
and modulus of Em• determine the stress in the matrix and in
each fiber whe11 the force Pis applied on the specimen.
p
i
i
p
Prob. 9-48
440
CHAPTER
9
AXIAL LOAD
9-49. The rigid bar is pinned at A and supported by two
aluminum rods, each having a diameter of 1 in .. a modulus of
elasticity £.1 =10(103) ksi, and yield stress of (uy)a1=40 ksi.
lf the bar is initially vertical, determine the displacement of the
end B when the force of 20 kip is applied.
9-53. The 2014-T6 aluminum rod AC is reinforced with the
firmly bonded A992 steel tube BC. If the assembly fits
snugly between the rigid supports so that there is no gap at C.
determine the support reactions when the axial force of
400 kN is applied. The assembly is attached at D.
9-50. The rigid bar is pinned at A and supported by two
aluminum rods, each having a diameter of 1 in., a modulus of
elasticity £ 31 =10(103) ksi, and yield stress of (uy)01 =40 ksi. If
the bar is i11itially vertical, determine the angle of tilt of the
bar when tJ1e 20-kip load is applied.
9-54. The 2014-T6 alwuinum rod AC is reinforced with the
firmly bonded A992 steel tube BC. When no load is applied
to the assembly, the gap between e nd C and the rigid
support is 0.5 mm. Determine the support reactions when
the axial force of 400 kN is applied.
20kip
B
I h
F
I
E
1.5h-ri1
2 ft
c.~
l-1.s --+I
D
A
400mm
400kN
-+-- - · ......'"-8
?
1 ft
h
A
_l
50
800mm
~2steel
a25 mmT2o14-T6 aluminum alloy
Probs. 9-49/50
Section~
9-51. The rigid bar is pinned at A and supported by two
aluminum rods. each having a diameter of l in. and a
modulus of elasticity £.1 =10(103) ksi. If the bar is initially
vertical. determine the displacement of the end 8 when the
force of 2 kip is applied.
*9-52. 111e rigid bar is pinned at A and supported by two
aluminum rods, each having a diameter of l in. and a
modulus of elasticity £.1 =10(103) ksi. If the bar is initially
vertical, determine the force in each rod when the 2-kip
load is applied.
Probs. 9-53/54
9-55. The three suspender bars are made of A992 steel
and have equal cross-sectional areas of 450 mm2 . Determine
the average normal stress in each bar if the rigid beam is
subjected to the loading shown.
B
I ft
F
E
I
f-1h-i!r.
t-+-1... 2 kip
c.
t
l -2h--J
1 rt
1 lft
I?
A
_I
Probs. 9-5V52
~
I
A
I
2m
D
0
c
80kN
SO kN
,__
I
I
B
E
0
!
ot"J
J-1 m--j--1 m-j-i m-j-t m-j
Prob. 9-55
F
9.6
9.6
THERMAL S TRESS
441
THERMAL STRESS
A change in temperature can cause a body to change its dimensions.
Generally, if the temperature increases, the body will expand, whereas if
the temperature decreases, it will contract.* Ordinarily this expansion or
contraction is linearly related to the temperature increase or decrease that
occurs. If this is the case, and the material is homogeneous and isotropic,
it has been found from experiment that the displacement of the end of a
member having a length L can be calculated using the formula
I Br = a6.TL I
(9-4)
Here
a
=
6. T
=
L =
or =
Most traffic bridges are designed wit h
expansion JOtn ts to accommodat e the
thermal move ment of the deck and thus
avoid any thermal stress.
a property of the material, referred to as the linear coefficient
of thermal expansion. The units measure strain per degree of
temperature. They are 1/°F (Fahrenheit) in the FPS system, and
1/°C (Celsius) or 1/K (Kelvin) in the SI system. Typical values
are given on the inside back cover.
the algebraic change in temperature of the member
the original length of the member
the algebraic change in the length of the member
The change in length of a statically determinate member can easily be
calculated using Eq. 9-4, since the member is free to expand or contract
when it undergoes a temperature change. However, for a statically
indeterminate member, these thermal displacements will be constrained
by the supports, thereby producing thermal stresses that must be
considered in design. Using the methods outlined in the previous sections,
it is possible to determine these thermal stresses, as illustrated in the
following examples.
*There are some materials, like Invar, an iron-nickel alloy, and scandium trifluoride, that
behave in the opposite way, but we will not consider these here.
j
Long extensions of ducts and pipes that carry
fluids are subjecte d to variations in
temperature that will cause them to expand
and contract. Expansion joints, such as the
one shown, are used to mitigate thermal
stress in the material.
442
I
CHAPTER
EXAMPLE
9
9.10 1
0.5 in.
1-1
D
A
AXIAL L OAD
I O.Sin.
TheA-36 steel bar shown in Fig. 9-17a is constrained to just fit between two
fixed supports when T1 = 60°F. 1f the temperature is raised to T2 = 120°F,
determine the average normal thermal stress developed in the bar.
SOLUTION
-
2 ft
Equilibrium. The free-body diagram of the bar is shown in Fig. 9- 17b.
Since there is no external load, the force at A is equal but opposite to the
force at B; that is,
+jIFy
B
(a)
F
i
=
O;
The problem is statically indeterminate since this force cannot be
determined from equilibrium.
Compatibility. Since oA/B = 0, the thermal displacement oT at A that
occurs, Fig. 9- 17c, is counteracted by the force F that is required to push
the bar oF back to its original position. The compatibility condition at A
becomes
(+j)
Load-Displacement.
relationships, we have
t
FL
0 = a6.TL - -
F
(b)
Applying the thermal and load-displacement
AE
Using the value of a on the inside back cover yields
F
=
=
=
a6.TAE
(6.60(10- 6)/°F](120°F - 60°F)(0.5 in.)2 (29(103) kip/in 2]
2.871 kip
Since F also represents the internal axial force within the bar, the
average normal compressive stress is thus
F
2.871 kip
.
.
a = - =
.
?
= 11.5 ks1
A
(0.5 m.)(c)
Fig. 9-17
Ans.
NOTE: The magnitude of F indicates that changes in temperature can
cause large reaction forces in statically indeterminate members.
9.6
EXAMPLE
9. ~
The rigid beam shown in Fig. 9-l&l is fixed to the top of the three posts
made of A992 steel and 2014-T6 aluminum. The posts each have a length of
250 mm when no load is applied to the beam, and the temperature is
T1 = 20"C. Determine the force supported by each post if the bar is
subjected to a uniform distributed load of 150 kN/m and the temperature
is raised to T2 = 80"C.
l-300 mm+300 mm-I
I l 1l l l l l l l Jtso
The free-body diagram of the beam is shown in Fig. 9-18b.
Moment equilibrium about the beam's center requires the forces in the
steel posts to be equal. Summing forces on the free-body diagram, we have
+ f "i.0, = 0;
2F,;1 + Fa1 - 90(103) N = 0
(1)
60mm-
Due to load, geometry, and material symmetry, the top
of each post is displaced by an equal amount. Hence,
( + l)
o.. = Bai
(2)
The final position of the top of each post is equal to its displacement
caused by the temperature increase, plus its displacement caused by the
internal axial compressive force, Fig. 9- 18c. Thus, for the steel and
aluminum post, we have
(+ l )
ost = - (o.,)r + (o.,)F
(+l)
Bai = - (o.1)r + (oa1)F
Applying Eq. 2 gives
-(o.,)r + (o.JF = - (oa1)r + (oai)F
Using Eqs. 9-2 and
properties on the inside back cover, we get
Load-Displacement.
9~
40mm- -
Aluminum
1- - - ,
•
i.-----,
I
'
F"
(b)
(c)
F.1 (0.250 m)
-[12(10 )/°C](80"C - 20"C)(0.250 m) + _ _ _::.:....:.._
_ ____:_ __
1T(0.020 m) 2 [200(10 9) N/m 2]
Fs1
= l.216Fa1 -
165.9(103)
~·~ i~·:~~l~\ N/m
(3)
To be consistent, all numerical data has been expressed in terms of
newtons, meters, and degrees Celsius. Solving Eqs. 1 and 3 simultaneously
yields
Ans.
F81 = - 16.4 kN Fa1 = 123 kN
The negative value for F 51 indicates that this force acts opposite to that
shown in Fig. 9i.-18b. In other words, the steel posts are in tension and the
aluminum post is in compression.
Steel
90kN
and the material
20°C)(0.250 m) + 1T(0.030
250mm
(a)
~
-[23(10~)/°C)(80"C -
T
<-
-40mm
Equilibrium.
Compatibility.
kN /m
I
Steel
SOLUTION
=
44 3
THERMAL STRESS
Fig. 9-18
2)
j
444
CHAPTER
EXAMPLE
9
AXIAL L OAD
9.12
-
~
l
150mm
J
A 2014-T6 aluminum tube having a cross-sectional area of 600 mm2 is used
as a sleeve for an A-36 steel bolt having a cross-sectional area of 400 mm2 ,
Fig. 9- 19a. When the temperature is T1 = l5°C, the nut holds the assembly
in a snug position such that the axial force in the bolt is negligible. If the
temperature increases to T2 = 80°C, determine the force in the bolt and
sleeve.
SOLUTION
Equilibrium. The free-body diagram of a top segment of the assembly
is shown in Fig. 9- 19b. The forces Fb and F, are produced since the sleeve
has a higher coefficient of thermal expansion than the bolt, and therefore
the sleeve will expand more when the temperature is increased. It is
required that
(a)
(1)
+ jIFy = O;
Compatibility. The temperature increase causes the sleeve and bolt to
expand (os)T and (ob)T, Fig. 9- 19c. However, the redundant forces Fb and
F, elongate the bolt and shorten the sleeve. Consequently, the end of the
assembly reaches a final position, which is not the same as its initial
position. Hence, the compatibility condition becomes
(+ !)
Load-Displacement. Applying Eqs. 9- 2 and 9-4, and using the
mechanical properties from the table on the inside back cover, we have
(b)
(12(10 - 6)/°C](80°C - l5°C)(0.150 m) +
Fb (0.150 m)
=
Initial
position
~
(ll,)r
(23(10 - 6) / °C](80°C - 15°C)(0.150 m)
~(llb)r
Fs (0.150 m)
ll
(llb)F
Final
position
~--~-
L
1
(ll,)F
Using Eq. 1 and solving gives
(c)
Fig. 9-19
Fs = Fb = 20.3 kN
Ans.
NOTE: Since linear elastic material behavior was assumed in this
analysis, the average normal stresses should be checked to make sure
that they do not exceed the proportional limits for the material.
9.6
THERMAL STRESS
445
PROBLEMS
*9-56. The C83400-red-brass rod AB and 2014-T6aluminum rod BC are joined at the collar B and fixed
connected at their ends. lf there is no load in the members
when T 1 = 50°F. determine the average normal stress in each
member when T2 = 120°F. Also. how far will the collar be
displaced? The cross-sectional area of each member is 1.75 in2•
9-59. The two cylindrical rod segments are fixed to the rigid
walls such that there is a gap of 0.01 in. between them when
T 1 = 60°F. What larger temperature T2 is required in order
to just close the gap? Each rod has a diameter of 1.25 in.
Detennine the average normal stress in each rod if
Ti = 300°F. Take a,1 = 13(1Q-6)/°F. £.1 = lO(lCP) ksi.
(uy)a1 = 40 ksi, acu = 9.4(1o-6)/°F. Ecu = 15(HP) ksi. and
(uy)cu = 50 ksi.
*~. The two cylindrical rod segments are fixed to the
rigid walls such that there is a gap of 0.01 in. between them
when T 1 = 60°F. Each rod has a diameter of 1.25 in.
Determine the average normal stress in each rod
if Ti = 400°F, and also calculate the new length of the
aluminum segment. Tuke a 81 = 13(1o-6)f°F, £ 31 = 10(103) ksi,
(uy) 31 = 40 ksi, acu = 9.4(1o-6)f°F, (uy)cu = 50 ksi, and
Ecu = 115(103) ksi.
Prob. 9-56
9-57. The assembly has the diameters and material
indicated. If it fits securely between its fixed supports
when the tempera ture is T1 =70°F, determine the average
normal stress in each material when the temperature
reaches T 2 = 11 0°F.
O.Olin .-1 ~
Co
~
r
12 in.
1
l-6
in.--1
Probs. 9-59/60
2014-T6 Aluminum
A
304 Stainless
C 86100 Bronze
steel
12 in.
~1.
D
The pipe is made of A992 steel and is connected to
the collars at A and B. When the temperature is 60°F, there is
no axial load in the pipe. If hot gas traveling through the pipe
causes its temperature to rise by ~ T = (40 + 15x)°F. where
x is in feet, determine the average normal stress in the pipe.
The inner diameter is 2 in .. the wall thickness is 0.15 in.
~2.
Prob. 9-57
9-58. The rod is made or A992 steel and has a diameter of
0.25 in. If the rod is 4 ft long when the springs are compressed
0.5 in.and the temperature of the rod is T = 40°F,determine
the force in the rod when its temperature is T = 160°F.
The bronze C86100 pipe bas an inner radius of
0.5 in. and a wall thickness of 0.2 in. lf the gas flowing
through it changes the temperature of the pipe uniformly
from TA = 200°F at A to T8 = 60°F at B, determine the
axial force it exerts on the walls. The pipe was fitted between
the walls when T = 60°F.
- - - 4 f t -- Prob. 9-58
Probs. 9~1/62
446
CHAPT ER
9
A XIAL L OAD
9-63. The 40-ft-long A-36 steel rails on a train track are laid
with a small gap between them to allow for thermal
expansion. Determine the required gap 8 so that the rails just
touch one another when the temperature is increased from
T1 = - 20°F to Ti = 90°F. Using this gap, what would be the
axial force in the rails if the temperature rises to T3 = 110°F?
The cross-sectional area of each rail is 5.10 in2 .
-11- s
s -11-
F
• •
• •
•
• •
40 ft
•
•
• •
-1
9-66. When the temperature is at 30°C, the A-36 steel pipe
fits snugly !between the two fuel tanks. When fuel flows
through the pipe, the temperatures at ends A and B rise to
130°C and &0°C, respectively. If the temperature drop along
the pipe is linear, determine the average normal stress
developed in the pipe. Assume each tank provides a rigid
support at A and B.
ISO~
lOmm
Section a - a
Prob. 9-63
*9-64. The device is used to measure a change in
temperature. Bars AB and CD are made of A-36 steel and
2014-T6 aluminum alloy, respectively. When the temperature
is at 75°F, ACE is in the horizontal position. Determine the
vertical displacement of the pointer at E when the
temperature rises to 150°F.
3in.=-i ~
r
A
0
0
c
E
1.5 in.
B
D
Prob. 9-64
9-65. The bar has a cross-sectional area A , length L ,
modulus of elasticity E, and coefficient of thermal expansion
a. The temperature of the bar changes uniformly along its
length from TA at A to T8 at B so that at any point x along
the bar T = TA + x(T8 - TA)/ L. Determine the force the
bar exerts on the rigid walls. Initially no axial force is in the
bar and the bar has a temperature of TA.
Prob.9-66
9-67. When the temperature is at 30°C, the A-36 steel pipe
fits snugly !between the two fuel tanks. When fuel flows
through the pipe, the temperatures at ends A and B rise to
130°C and &0°C, respectively. If the temperature drop along
the pipe is linear, determine the average normal stress
developed in the pipe. Assume the walls of each tank act as
a spring, each having a stiffness of k = 900 MN/m.
*9-68. When the temperature is at 30°C, the A-36 steel
pipe fits snugly between the two fuel tanks. When fuel flows
through the pipe, it causes the temperature to vary along
the pipe as T = (~x2 - 20x + 120)°C, where x is in meters.
Determine the normal stress developed in the pipe. Assume
each tank provides a rigid support at A and B.
150m~
lOmm~
Section a - a
- x-J
A
B
a
Prob. 9-65
Probs. 9...(,7/68
B
9.6
9-69. The SO-mm-diameter cylinder is made from
Am 1004-T61 magnesium and is placed in the clamp when
the temperature is Ti = 20° C. If the 304-stainless-steel
carriage bolts of the clamp each have a diameter of 10 mm,
and they hold the cylinder snug with negligible force against
the rigid jaws, determine the force in the cylinder when the
temperature rises to T2 = 1300C.
9-70. The SO-mm-diameter cylinder is made from
Am 1004-T61 magnesium and is placed in the clamp when
the temperature is Ti = 1S0 C. If the two 304-stainless-steel
carriage bolts of the clamp each have a diameter of 10 mm,
and they hold the cylinder snug with negligible force against
the rigid jaws, determine the temperature at which the
average normal stress in eithe r the magnesium or the steel
first becomes 12 MPa.
THERMAL STRESS
447
*9-72. The cylinder CD of the assembly is heated from
T 1 = 30°C to T2 = 180°C using electrical resistance. At the
lower temperature T1 the gap between C and the rigid bar is
0.7 mm. Determine the force in rods AB and EF caused by
the increase in temperature. Rods AB and EF are made of
steel, and each has a cross-sectional area of 12S mm2• CD is
made of aluminum and has a cross-sectional area of 37S mm2 .
£., = 200GPa,Eai = 70GPa.and aa1 = 23(1CJf>)/°C.
9-73. The cylinder CD of the assembly is heated from
T1 = 30°C to T2 = 180°C using electrical resistance. Also. the
two end rods AB and EF are heated from T1 = 30°C to
T2 = S0°C. At the lower temperature Ti the gap between C
and the rigid bar is 0.7 mm. Determine the force in rods AB
and EF caused by the increase in temperature. Rods AB
and EF are made of steel, and each has a cross-sectional
area of 12S mm 2 . CD is made of aluminum and has a crosssectional area of 37S mm2 . £,, = 200 GPa, £ 01 = 70 GPa,
asi = 12(1~)/°C, and a 01 = 23(1CJf>)j°C.
0.7mm
F
150mm
IOOmm
l.~
300mm
A
Probs. 9-69no
9-71. The wires AB and AC are made of steel. and wire
AD is made of copper. Before the lSO-lb force is applied,
AB and AC are each 60 in. long and AD is 40 in. long. If the
temperature is increased by SO"F. determine the force in
each wire needed to support the load. Each wire has a crosssectional area of 0.0123 in2. Take £., = 29(10 3) ksi,
Ecu = 17(10 3) ksi, a,1 = 8(10-6)/°F, acu = 9.(i()(l0-6)/°F.
Probs. 9-71f13
9-74. The metal strap has a thickness / and width 111 and is
subjected to a temperature gradient Ti to Tz (Ti < Tz).
This causes the modulus of elasticity for the material to vary
linearly from Ei at the top to a smaller amount £ 2 at the
bottom. As a result. for any vertical position y. measured
from the top surface,£ = ((£2 - Ei) /wjy + Ei. Determine
the position d where the axial force P must be applied so
that the bar stretches uniformly over its cross section.
c
B
1501b
Prob. 9-71
Prob. 9-74
448
9
CHAPTER
AXIAL L OAD
CONCEPTUAL PROBLEMS
C9-L The concrete footing A was poured when this
column was put in place. Later the rest of the foundation
stab was poured. Can you explain why the 45° cracks
occured at each corner? Can you think of a better design
that would avoid such cracks?
C9- 2. The row of bricks, along with mortar and an internal
steel reinforcing rod, was intended to serve as a lintel beam
to support the bricks above this ventilation opening on an
exterior wall of a building. Explain what may have caused
the bricks to fail in the manner shown.
Prob. C9-1
Prob. C9-2
CHAPTER REVIEW
When a loading is applied at a point on a body, it tends
to create a stress distribution within the body that
becomes more uniformly distributed at regions
removed from the point of application of the load.
This is called Saint-Venant's principle.
N
1-U a\'g =
The relative displacement at the end of an axially
loaded member relative to the other end is
determined from
ii =
1
L
0
N(x)dx
AE
1----
x--- 11- dx
1----11--+ - -+ i
Pi .....
AN
_,
:----L.
----L
P,
~I
CHAPTER REVIEW
449
U a series of co ncentrated external axial forces are
applied to a member and A£ is also piecewise constant,
then
NL
6 =}: AE
For application. it is necessary to use a sign convention
for the internal load N and displacement 8. We consider
tension and elongation as positive values. Also, the
material must not yield. but rather it must remain linear
elastic.
Superposition of load and displacement is possible
provided tbe material remains linear elastic and no
significant cha nges in the geometry of the member
occur after loadi ng.
The reactions on a statically indeterminate bar can be
determined using the equilibrium equations
and compatibility conditions that specify tbe
displacement at the supports. These displacements are
related to the loads using a load--<lisplacement
relationship such as 8 = NL/ A£.
A change in temperature can cause a member made of
homogeneous isotropic material to change its length by
8 =at.TL
U the member is confined. this change will produce
thermal stress in the member.
Holes and sharp transi ti ons at a cross section will
create stress conce ntrations. For the design of a
member made of brittle material one obtains the
stress concentration factor K from a graph, wh ich has
been determined from experiment. This va lue is then
multiplied by the average stress to obtain the
maximum stress at the cross section.
P1
•
LI_
_
P_2_.,.~
_-_-:_ _ _ _.._--~~-P_3_
__,[1 •
:1--~~~~L~~~~~41~sl
P4
450
CHAPTER
9
AXIAL L OAD
REVIEW PROBLEMS
R9-1. The assembly consists of two A992 steel bolts AB
and EF and an 6061-T6 aluminum rod CD. When the
temperature is at 30° C, the gap between the rod and rigid
member AE is 0.1 mm. Determine the normal stress
developed in the bolts and the rod if the temperature rises
to 130° C. Assume BF is also rigid.
R9-3. The rods each have the same 25-mm diameter and
600-mm length. If they are made of A992 steel, determine
the forces developed in each rod when the temperature
increases by 50° c.
c
c
25mm
25mm
300mm
400mm
SO
B
1 - -600 mm - --'
0
B
mm
D
A
600mm
~--~----'
v~
Prob. R9-1
Prob. R9- 3
R9-2. The assembly shown consists of two A992 steel bolts
AB and EF and an 6061-T6 aluminum rod CD. When the
temperature is at 30° C, the gap between the rod and rigid
member AE is 0.1 mm. Determine the highest temperature
to which the assembly can be raised without causing yielding
either in the rod or the bolts. Assume BF is also rigid.
c
25mm
*R9-4. Two A992 steel pipes, each having a cross-sectional
area of 0.32 in2 , are screwed together using a union at B.
Originally the assembly is adjusted so that no load is on the
pipe. If the union is then tightened so that its screw, having a
lead of 0.15 in., undergoes two full turns, determine the
average normal stress developed in the pipe. Assume that
the union and couplings at A and C are rigid. Neglect the
size of the union. Nore: The lead would cause the pipe, when
unloaded, to shorten 0.15 in. when the union is rotated one
revolution.
25mm
300mm
400mm
SO
mm
D
- - - 3ft - - -
Prob. R9-2
Prob. R9-4
451
REVIEW PROBLEMS
R9-5. The 2014-T6 aluminum rod has a diameter of
0.5 in. and is lightly attached to the rigid supports at A and B
when T1 = 70"F. Lf the temperature becomes T2 =- 10°F, and
an axial force of P = 16 lb is applied to the rigid collar as
shown. determine the reactions at the rigid supports A and B.
R9- 7. The rigid link is supported by a pin at A and
two A-36 steel wires. each having an unstretched length of
12 in. and cross-sectional area of 0.0125 in2 . Determine the
force developed in the wires when the link supports the
vertical load of 350 lb.
- - - - 1 2 in.-- - - l
1
c
T
5 in.
B
A
+~
B~
Pf2.
Pf2.
4
- S in.
in.
I
Prob. R9-5
-
6in.-I
3501b
Prob. R9-7
R9-6. The 2014-T6 aluminum rod has a diameter of
0.5 in. and is lightly allached to the rigid supports at A and B
when T 1 = 70°F. Determine the force P that must be applied
to the collar so that. when T = 0°F, the reaction at Bis zero.
B
A
Pf2.
Pf2. -
*R9-8. The joint is made Crom three A992 steel plates that
are bonded together at their seams. Determine the
displacement of end A with respect to end B when the joint
is subjected to the axial loads. Each plate has a thickness of
5mm.
5 in. -
J
46kN
I 1- - - 8 in. Prob. R9-6
~~
23kN
;:
I B 23 kN
--:rl
Prob. R9-8
CHAPTER
10
(©Jill Fromer/Getty Images)
The torsiona l stress and angle of twist of this soil auger depend upon the output
of the machine turning the bit as well as the resistance of the soi l in contact with
the shaft.
TORSION
CHAPTER OBJECTIVES
•
To determine the torsional stress and deformation of an elastic
circular shaft.
•
To determine the support reactions on a statically indeterminate,
torsionally loaded shaft when these reactions cannot be
determined solely from the moment equilibrium equation.
10.1
TORSIONAL DEFORMATION OF A
CIRCULAR SHAFT
Torque is a moment that tends to twist a member about its longitudinal
axis. Its effect is of primary concern in the design of drive shafts used in
vehicles and machinery, and for this reason it is important to be able to
determine the stress and the deformation that occurs in a shaft wh.en it is
subjected to torsional loads.
We can physically illustrate what happens when a torque is applied to
a circular shaft by considering the shaft to be made of a highly deformable
material such as rubber. When the torque is applied, the longitudinal grid
lines originally marked on the shaft, Fig. 10- la, tend to distort into a
helix, Fig. 10- lb, that intersects the circles at equal angles. Also, all the
cross sections of the shaft will remain flat - that is, they do not warp or
bulge in or out- and radial lines remain straight and rotate during this
deformation. Provided the angle of twist is small, then the length of
the shaft and its radius will remain practically unchanged.
453
454
CHAPTER
10
TORSION
If the shaft is fixed at one end and a torque is applied to its other end,
then the dark green shaded p[ane in Fig. 10- 2a will distort into a skewed
form as shown. Here a radial line located on the cross section at a
distance x from the fixed end of the shaft will rotate through an angle
<f>(x). This angle is called the angle of twist. It depends on the position x
and will vary along the shaft as shown.
In order to understand how this distortion strains the material, we will
now isolate a small disk element located at x from the end of the shaft,
Fig. 10-2b. Due to the deformation, the front and rear faces of the
element will undergo rotation-the back face by <f>(x), and the front face
by <f>(x) + d<f>. As a result, the difference in these rotations,d<f>, causes the
element to be subjected to a shear strain, y (see Fig. 8- 24b ).
Before deformation
(a)
Circles remain
circular
T
Longitudinal
lines become
twisted
T
Radial lines
remain straight
Notice the deformation of the rectangular
element when this rubber bar is subjected
to a torque.
After deformation
(b)
Fig.10-1
10.1
TORSIONAL DEFORMATION OF A CIRCULAR SHAFT
Deformed
plane
Undeformed
plane
The angle of twist </l(x) increases asx increases.
The shear strain at points on
the cross section increases linearly
with p, i.e., Y = (p/chmax·
(b)
(a)
Fig. 10-2
This angle (or shear strain) can be related to the angle def> by noting that
the length of the red arc in Fig. 10-2b is
p def>= dx y
or
'Y
=
def>
dx
p-
(10-1)
Since dx and def> are the same for all elements, then d<f>/dx is constant
over the cross section, and Eq. 10-1 states that the magnitude of the
shear strain varies only with its radial distance p from the axis of the
shaft. Since d<f>/dx = y/ p = 'Ymax/c, then
(10-2)
In other words, the shear strain within the shaft varies linearly along any
radial line, from zero at the axis of the shaft to a maximum 'Yma:x at its
outer boundary, Fig. 10-2b.
455
456
CHAPTER
10
TORSION
10.2 THE TORSION FORMULA
When an external torque is applied to a shaft, it creates a corresponding
internal torque within the shaft. In this section, we will develop an
equation that relates this internal torque to the shear stress distribution
acting on the cross section of the shaft.
If the material is linear elastic, then Hooke's law applies, T = Gy , or
Tmax = G'Ymax, and consequently a linear variation in shear strain, as
noted in the previous section, leads to a corresponding linear variation
in shear stress along any radial line. Hence, T will vary from zero at the
shaft's longitudinal axis to a maximum value, Tmax, at its outer surface,
Fig. 10-3. Therefore, similar to Eq. 10-2, we can write
T
=
(~ )Tmax
T
Shear stress varies linearly along
each radial line of the cross section.
Fig.10.-3
(10-3)
10.2
THE T ORSION FORMULA
457
Since each e le me nt of area dA, located at p, is subjected to a force of
dF = T dA, Fig. 10-3, the torque produced by this force is then
dT = p(r dA). Fo r the e ntire cross section we have
(10-4)
However, Tmax/c is constant, and so
T
= Tmax
r
p2
C }A
dA
(10-5)
The integral represents the polar moment of inertia of the shaft's
cross-sectional area about the shaft's longitudinal axis. On the next page
we will calculate its value, but here we will symbolize its value as J. As a
result, the above equation can be rearranged and written in a more
compact form , name ly,
(10-6)
Here
Tmax
=
T
=
J
= the polar moment of inertia of the cross-sectional area
c
=
the maximum shear stress in the shaft, which occurs at its outer
surface
the resultant internal torque acting at the cross section. Its value
is determined from the method of sections and the equation of
mome nt e quilibrium applied about the shaft's longitudinal axis
the outer radius of the shaft
If Eq.10-6 is substituted in Eq.10-3, the shear stress at the inte rmediate
distance p on the cross section can be determined.
(10-7)
Either of the above two equations is often referred to as the torsion
formula . Recall that it is used only if the shaft has a circular cross section
and t he mate rial is ho mogeneous and behaves in a linear e lastic manner,
since the derivation of Eq. 10-3 is based on Hooke's law.
The sha ft auached to the center of this
wheel is subj ected to a torque, and the
maximum stress it creates must be resisted
by the shaft lo prevent failure.
458
CHAPTER
10
TORSION
Fig. 10-4
Polar Moment of Inertia. If the shaft has a solid circular cross
section, the polar moment of inertia J can be determined using an area
element in the form of a differential ring or annulus having a thickness dp
and circumference 2TTp, Fig. 10-4. For this ring, dA = 2TTp dp, and so
J = lp2 dA = 1cp2(27rp dp)
=
271' focp3dp
IJ
=
; c4
2~~)p4 :
I
(10-8)
Solid Section
Note that J is always positive. Common units used for its measurement
are mm4 or in4.
If a shaft has a tubular cross section, with inner radius C; and outer
radius c0 , Fig. 10-5, then from Eq. 10-8 we can determine its polar
moment of inertia by subtracting J for a shaft of radius C; from that
determined for a shaft of radius c0 . The result is
I1 =
T(c~ - c[) I
Tube
Fig.10-5
(10-9)
10.2
THE T ORSION FORMULA
459
/
T
Shear stress varies linearly along
each radial line of the cross section.
(b}
(a)
Fig. 10-6
Shear Stress Distribution. If an element of material on the cross
section of the shaft or tube is isolated, then due to the complementary
property of shear, equal shear stresses must also act on four of its adjacent
faces, as shown in Fig. 10-6a. As a result, the internal torque T develops
a linear distribution of shear stress along each radial line in the plane
of the cross-sectional area, and also an associated shear-stress
distribution is developed along an axial plane, Fig.10-6b. It is interesting
to note that because of this axial distribution of shear stress, shafts made
of wood tend to split along the axial plane when subjected to excessive
torque, Fig. 10-7. This is because wood is an anisotropic material, whereby
its shear resistance paraJlel to its grains or fibers, directed along the axis
of the shaft, is much less than its resistance perpendicular to the fibers
within the plane of the cross section.
7
T
T
Failure of a wooden shaft due to torsion.
Fig. 10-7
The tubular drive shaft for a truck was
subjected to an excessive torque, resulting
in failure caused by yielding of the material.
Engineers deliberately design drive shafts
to fail before torsional damage can occur
10 parts of the engine or transmission.
460
CHAPTER
10
TORSION
IMPORTANT POINTS
• When a shaft having a circular cross section is subjected to a torque, the cross section remains plane
while radial lines rotate. This causes a shear strain within the material that varies linearly along any radial
line, from zero at the axis of the shaft to a maximum at its outer boundary.
• For linear elastic homogeneous material, the shear stress along any radial line of the shaft also varies linearly,
from zero at its axis to a maximum at its outer boundary. This maximum shear stress must not exceed the
proportional limit.
• Due to the complementary property of shear, the linear shear stress distribution within the plane of the cross
section is also distributed along an adjacent axial plane of the shaft.
• The torsion formula is based on the requirement that the resultant torque on the cross section is equal
to the torque produced by the shear stress distribution about the longitudinal axis of the shaft. It is
required that the shaft or tube have a circular cross section and that it is made of homogeneous material
which has linear elastic behavior.
PROCEDURE FOR ANALYSIS
The torsion formula can be applied using the following procedure.
Internal Torque.
• Section the shaft perpendicular to its axis at the point where the shear stress is to be determined, and use
the necessary free-body diagram and equations of equilibrium to obtain the internal torque at the section.
Section Property.
• Calculate the polar moment of inertia of the cross-sectional area. For a solid section of radius c,J =
and for a tube of outer radius c0 and inner radius C;, J = 1T ( c~ - c[) /2.
4
1TC /2,
Shear Stress.
• Specify the radial distance p, measured from the center of the cross section to the point where the shear
stress is to be found. Then apply the torsion formula T = Tp/J, or if the maximum shear stress is to be
determined use Tmax = Tc/ J. When substituting the data, make sure to use a consistent set of units.
• The shear stress acts on the cross section in a direction that is always perpendicular to p. The force it
creates must contribute a torque about the axis of the shaft that is in the same direction as the internal
resultant torque T acting on the section. Once this direction is established, a volume element located at
the point where T is determined can be isolated, and the direction of T acting on the remaining three
adjacent faces of the element can be shown.
10.2
EXAMPLE
THE T ORSION FORMULA
4 61
10.1
The solid shaft and tube shown in Fig. 10--8 are made of a material having an
allowable shear stress of 75 MPa. D etermine rhe maximum torque rhat can
be applied to each cross section, and show rhe stress acting on a small
element of material at point A of the shaft, and points B and C of the rube.
SOLUTION
Section Properties. The polar moments of inertia for the solid and
tubular shafts are
3 4
4
4
1' = ~
2c = ~
2 (0. I m) = 0 .1571(10- ) m
(]j
75 MPa A
J, = ; (c~ - cf) = ; [ (0.1 m) 4
-
(0.075 m)4 ] = 0.1074(10- 3 ) m4
Shear Stress. The maximum torque in each case is
T,(0.1 m)
)
75 ( 106 N / m2 - -----'-----'-0.1571 ( 10- 3 ) m4
T=118kN·m
'
Ans.
"")
,
T,(0.1 m)
75 ( lu- N/m- = -0 .-1 07
--'--'
4 -( 1-0--3-'-)- m
-4
T, = 80.5 kN · m
Also, the shear stress at the inner radius of the rube is
(7:)
I
I
=
~
Ans.
80.5 ( Ia3) N · m (0.075 m)
= 56 2 MPa
0.1074(10- 3) m4
•
These results are shown acting on small elements in Fig. 10--8. Notice
how rhe shear stress o n the front (shaded) face of the element contributes
to the torque. As a consequence, shear stress components act on the
other three faces. No shear stress acts on the outer surface of the shaft or
tube or on the inne r surface of the tube because it must be stress free.
56.2MPa
Fig. 10--8
462
CHAPTER
EXAMPLE
10
TORSION
10.2
-
-
The 1.5-in.-diameter shaft shown in Fig. 10-9a is supported by two
bearings and is subjected to three torques. Determine the shear stress
developed at points A and B, located at section a- a of the shaft, Fig.10-9c.
30 kip·in.
~
(a)
x
(b)
~18.9ksi
SOLUTION
~2.5
kip·in.
Internal Torque. Since the bearing reactions do not
offer resistance to shaft rotation, the applied torques
sat isfy moment equilibrium about the shaft's axis.
The internal torque at section a-a will be determined from
the free-body diagram of the left segment, Fig.10-9b. We have
~B
3.77 ksi
x
(c)
Fig.10-9
IM,
=
O;
42.5 kip· in. - 30 kip· in. - T = 0
Section Property.
T
=
12.5 kip· in.
The polar moment of inertia for the shaft is
4
4
J = ; (0.75 in. ) = 0.497 in
Shear Stress.
Since point A is at p = c = 0.75 in.,
Tc
TA =
J
( 12.5 kip· in.) ( 0.75 in.)
=
18.9 ksi
Ans.
(12.5kip·in. )(0.15in. )
( 0.4
in )
= 3.77 ksi
97 4
Ans.
( 0.4
97
in4 )
=
Likewise for point B, at p = 0.15 in., we have
Tp
T8 =
J
=
NOTE: The directions of these stresses on each element at A and B,
Fig. 10- 9c, are established on the planes of each of these elements, so
that they match the required clockwise torque.
10.2
I
EXAMPLE
THE TORSION FORMULA
463
10.3 1
The pipe shown in Fig. 10-lOa has an inner radius of 40 mm and an outer
radius of 50 mm. If its end is tightened against the support at A using the
torque wrench, determine the shear stress developed in the material at
the inner and outer walls along the central portion of the pipe.
SON
~lzoomm
SOLUTION
(a)
A section is taken at the intermediate location C
along the pipe's axis, Fig. 10-lOb. The only unknown at the section is the
internal torque T . We require
Internal Torque.
IMx = O·,
80 N ( 0.3 m) + 80 N ( 0.2 m) - T
T
=
=
0
40N · m
Section Property. The polar moment of inertia for the pipe's
cross-sectional area is
J = ; [ ( 0.05 m) 4 - ( 0.04 m ) 4 ] = 5.796(10- 6 ) m4
Shear Stress. For any point lying on the outside surface of the pipe,
p = c0 = 0.05 m, we have
Tc0
r.0 = -
J
=
40 N · m(0.05 m)
5.796(10- 6 ) m4
=
0.345 MPa
And for any point located on the inside surface, p
T.· =
'
Tc;
-
J
=
40N · m(0.04m)
5.796(10- 6 ) m4
=
= C; =
0.04 m , and so
0.276 MPa
~
Ans. Stress free
The results are shown on two small elements in Fig.10-lOc.
D and the inner face of E are in stress-free
regions, no shear stress can exist on these faces or on the other
corresponding faces of the elements.
NOTE: Since the top face of
~MPa
Ans.
inside
(c)
Fig. 10-10
464
CHAPTER
10
TORSION
10. 3
POWER TRANSMISSION
Shafts and tubes having circular cross sections are often used to transmit
power developed by a machine. When used for this purpose, they are
subjected to a torque that depends on both the power generated by the
machine and the angular speed of the shaft. Po wer is defined as the work
performed per unit of time. Also, the work transmitted by a rotating shaft
equals the torque applied times the angle of rotation. Therefore, if during
an instant of time dt an applied torque T causes the shaft to rotate dO, then
the work done is TdO and the instantaneous power is
P = TdO
dt
Since the shaft's angular velocity is w
The belt drive transmits the torque
developed by an electric motor to
the shaft at A. The stress developed
in the shaft depends upon the power
transmitted by the motor and the
rate ofrotation of the shaft. P = Tw.
=
d() / dt, then the power is
(10- 10)
In the SI system, power is expressed in watts when torque is measured
in newton-meters ( N · m) and w is in radians per second ( rad/s)
( 1 W = 1 N · m/s) . In the FPS system, the basic units of power are
foot-pounds per second (ft· lb/s); however, horsepower (hp) is often
used in engineering practice, where
1 hp
=
550 ft· lb/s
For machinery, the frequency of a shaft's rotation,/, is often reported.
This is a measure of the number of revolutions or "cycles" the shaft
makes per second and is expressed in hertz ( 1 Hz = 1 cycle/s). Since
1 cycle = 21T rad, then w = 21Tf, and so the above equation for power
can also be written as
(10- 11)
Shaft Design.
When the power transmitted by a shaft and its
frequency of rotation are known, the torque developed in the shaft can
be determined from Eq. 10-11, that is, T = P/21Tf Knowing T and the
allowable shear stress for the material, TaJJow, we can then determine the
size of the shaft's cross section using the torsion formula. Specifically, the
design or geometric parameter J/ c becomes
J
T
C
Tallow
-=--
(10- 12)
For a solid shaft, J = ( 1T /2) c 4 , and thus, upon substitution, a unique value
for the shaft's radius c can be determined. If the shaft is tubular, so that
J = ( 1T /2) ( c~ - c{), design permits a wide range of possibilities for the
solution. This is because an arbitrary choice can be made for either c0 or c;
and the other radius can then be determined from Eq. 10-12.
10.3
EXAMPLE
10.4
A solid steel shaft AB, shown in Fig. 10-11, is to be used to transmit 5 hp
from the motor M to which it is anached. If the shaft rotates at w = 175 rpm
and the steel has an allowable shear stress of TaJJow = 14.5 ksi, determine
the required diameter of the shaft to the nearest in.
k
Fig. 10-11
SOLUTION
The torque on the shaft is determined from Eq. 10-10, that is, P = Tw.
Expressing Pin foot-pounds per second and win radians/second, we have
P = 5 hp (
w
=
550 ft· lb/s)
= 2750ft · lb/s
1 hp
175 .rev ( 2'1T rad )( 1 min )
mm
1 rev
60 s
= 1833 rad/s
Thus,
P
=
2750 ft· lb/s
Tw;
=
T( 18.33 rad/s)
T = 150.1 ft· lb
Applying Eq.10-12,
J
C
c
=(
13
2T ) 1
'1T1i1llow
T
'1T c4
=-- = - -
2
C
'Tallow
= (2( 150.1 ft· lb) ( 12 in./ft) )1/3
'1T (
14 500 lb / in2 )
c = 0.429 in.
Since 2c = 0.858 in., select a shaft having a diameter of
d
= ~ in. = 0.875 in.
Ans.
POWER TRANSMISSION
465
466
C HAPT ER
10
TORSION
PRELIMINARY PROBLEMS
Pl0--1. Determine the internal torque at each section and
show the shear stress on differential volume elements
located at A , B, C, and D.
A
Pl0--3. The solid and hollow shafts are each subjected to
the torque T. In each case, sketch the shear stress distribution
along the two radial lines.
c
8
/ ;00N·m
0
300N·m
Prob. Pl0--1
Pl0--2. Determine the internal torque at each section and
show the shear stress on differential volume elements
located at A , B, C, and D.
~ 400N·m
Prob. Pl0--3
Pl0-4. The motor delivers 10 hp to the shaft. If it rotates
at 1200 rpm, detemine the torque produced by the motor.
~-
A
B
~
c
600N·m
~
D
Prob. Pl0--2
Prob. Pl0-4
10.3
POWER T RANSMISSION
467
FUNDAMENTAL PROBLEMS
• 10-1. The solid circular shaft is subjected to an internal
torque of T = 5 kN · m. Determine the shear stress at
points A and 8. Represent each state of stress on a volume
element.
.10-3. The shaft is hollow from A to Band solid from B to C.
Determine the maximum shear stress in the shaft. The shaft
has an outer diameter of 80 mm. and the thickness of the wall
of the hollow segment is 10 mm.
2 kN·m
Prob. F 10-1
Proh. Fl0-3
• 11.)....2 The hollow circular shaft is subjected to an internal
torque of T = 10 kN · m. Determine the shear stress at
points A and 8. Represent each state of stress on a volume
element.
60 111111
Pre . F 0-2
110-4. Determine the maximum shear stress m the
40-mm-diameter shaft.
6 kN
Proh. Fl0-4
468
CHAPTER
10
TORSION
Fl0-5. Determine the maximum shear stress in the shaft
at section a- a.
Fl0-7. The solid SO-mm-diameter shaft is subjected to the
torques applied to the gears. Determine the absolute
maximum shear stress in the shaft.
a
600N·m
~m~Omm1sooN·mc0
B
4omzy
soom{ c I
400n~
Section a-a
600N·m
"' "'
500mm
Prob.Fl0-5
Prob. Fl0-7
Fl0-6. Determine the shear stress at point A on the surface
of the shaft. Represent the state of stress on a volume
element at this point. The shaft has a radius of 40 mm.
Fl0-8. The gear motor can develop 3 hp when it turns at
150 rev /min. If the allowable shear stress for the shaft is
r allow = 12 ksi, determine the smallest diameter of the shaft
to the nearest kin. that can be used.
Prob.Fl0-6
Prob. Fl0-8
10.3
POWER T RANSMISSION
469
PROBLEMS
10-1. The solid shaft of radius r is subjected to a torque T.
Determine the radius r' of the inner core of the shaft that
resists one-half of the applied torque (T/2). Solve the
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
*10-4. The copper pipe has an outer diameter of 40 mm
and an inner diameter of 37 mm. If it is tightly secured to
the wall and three torques are applied to it, determine the
absolute maximum shear stress developed in the pipe.
10-2. The solid shaft of radius r is subjected to a torque T .
Determine the radius r' of the inner core of the shaft that
resists one-quarter of the applied torque (T/4). Solve the
problem two ways: (a) by using the torsion formula, (b) by
finding the resultant of the shear-stress distribution.
~
~y30N·m
~20Nm
SON·m
Prob. 10-4
T
Probs. 10-1/2
10-3. A shaft is made of an aluminum alloy having an
allowable shear stress of 'T a11ow = 100 MPa. If the diameter of
the shaft is 100 mm. determine the maximum torque T that
can be transmitted. What would be the maximum torque T ' if
a 75-mm-diameter hole were bored through the shaft? Sketch
the shear-stress distribution along a radial line in each case.
Prob. 10-3
10-5. The copper pipe has an outer diameter of 2.50 in.
and an inner diameter of 2.30 in. If it is tightly secured to the
wall and three torques are applied to it, determine the shear
stress developed at points A and 8. These points lie on the
pipe's outer surface. Sketch the shear stress on volume
elements located at A and 8.
Prob. 10-5
470
CHAPTER
10
TORSION
10-6. The solid aluminum shaft has a diameter of 50 mm
and an allowable shear stress of r allow = 60 MPa. Determine
the largest torque T 1 that can be applied to the shaft if it is
also subjected to the other torsional loadings. It is required
that T 1 act in the direction shown. Also, determine the
maximum shear stress within regions CD and DE.
10-9. The solid shaft is fixed to the support at C and
subjected to the torsional loadings. Determine the shear
stress at points A and Bon the surface, and sketch the shear
stress on volume elements located at these points.
10-7. The solid aluminum shaft has a diameter of 50 mm.
Determine the absolute maximum shear stress in the shaft and
sketch the shear-stress distribution along a radial line of the
shaft where the shear stress is maximum. Set T 1 = 2000 N · m.
c
"(~
J5m~20mm
~V-
300N·m
800 N·m
300 N·m
Prob.10-9
900N·m
Probs. 10-6n
*10-8. The solid 30-mm-diameter shaft is used to transmit
the torques applied to the gears. Determine the absolute
maximum shear stress in the shaft.
10-10. The link acts as part of the elevator control for a
small airplane. If the attached aluminum tube has an inner
diameter of 25 mm and a wall thickness of 5 mm, determine
the maximum shear stress in the tube when the cable force
of 600 N is applied to the cables. Also, sketch the
shear-stress distribution over the cross section.
300 N·m
600N
75L
_l
75L
500mm
_l
600N
Prob.10-8
Prob.10-10
10.3
10-11. The assembly consists of two sections of galvanized
steel pipe connected together using a reducing coupling at B.
The smaller pipe has an outer diameter of 0.75 in. and an inner
diameter of 0.68 in., whereas the larger pipe has an outer
diameter of 1 in. and an inner diameter of 0.86 in. If the pipe is
tightly secured to the wall at C, determine the maximum shear
stress in each section of the pipe when the couple is applied to
the handles of the wrench .
471
POWER TRANSMISSION
10-14. A steel tube having an outer diameter of 2.5 in. is
used to transmit 9 hp when turning at 27 rev /min. Determine
the inner diameter d of the tube to the nearest in. if the
allowable shear stress is r allow = 10 ksi.
l
., p
'8h1i
Prob.10-14
15 lb
Prob.10-11
*10-12. The shaft has an outer diameter of 100 mm and an
inner diameter of 80 mm. If it is subjected to the three
torques, determine the absolute maximum shear stress in
the shaft. The smooth bearings A and B do not resist torque.
10-13. The shaft has an outer diameter of 100 mm and an
inner diameter of 80 mm. If it is subjected to the three
torques, plot the shear stress distribution along a radial line
for the cross section within region CD of the shaft. The
smooth bearings at A and B do not resist torque.
10-15. If the gears are subjected to the torques shown,
determine the maximum shear stress in the segments AB and
BC of the A-36 steel shaft. The shaft has a diameter of 40 mm.
*10-16. If the gears are subjected to the torques shown,
determine the required diameter of the A-36 steel shaft to
the nearest mm if r allow = 60 MPa.
E
200N·m
c
SkN·m
Probs. 10-12113
Probs. 10-15/16
472
CHAPTER
10
TORSION
10-17. The rod has a diameter of 1 in. and a weight of
10 lb/ft. Determine the maximum torsional stress in the rod
at a section located at A due to the rod's weight.
10-18. The rod has a diameter of 1 in. and a weight of
15 lb/ft. Determine the maximum torsional stress in the rod
at a section located at B due to the rod's weight.
10-21. The 60-mrn-diameter solid shaft is subjected to the
distributed and concentrated torsional loadings shown.
Determine the shear stress at points A and B, and sketch
the shear stress on volume elements located at these points.
10-22. The 60-mrn-diameter solid shaft is subjected to the
distributed and concentrated torsional loadings shown.
Determine the absolute maximum and minimum shear
stresses on the shaft's surface, and specify their locations,
measured from the fixed end C.
10-23. The solid shaft is subjected to the distributed and
concentrated torsional loadings shown. Determine the
required diameter d of the shaft if the allowable shear stress
for the material is Tallow = 1.6 MPa.
Probs. 10-17/18
10-19. The copper pipe has an outer diameter of 3 in. and
an inner diameter of2.5 in. If it is tightly secured to the wall
at C and a uniformly distributed torque is applied to it as
shown, determine the shear stress at points A and B. These
points lie on the pipe's outer surface. Sketch the shear stress
on volume elements located at A and B.
*10-20. The copper pipe has an outer diameter of 3 in. and
an inner diameter of 2.50 in. If it is tightly secured to the
wall at C and it is subjected to the uniformly distributed
torque along its entire length, determine the absolute
maximum shear stress in the pipe. Discuss the validity of
this result.
c
Probs. 10-21122123
*10-24. The 60-mm-diameter solid shaft is subjected to
the distributed and concentrated torsional loadings shown.
Determine the absolute maximum and minimum shear
stresses in the shaft's surface and specify their locations,
measured from the free end.
10-25. The solid shaft is subjected to the distributed and
concentrated torsional loadings shown. Determine the
required diameter d of the shaft if the allowable shear stress
for the material is Tallow = 60 MPa.
B
150 lb·ft/ft
c
400 N·m
4 kN ·m/m ,
~"..._.."'
,~osm
800N~0.5m
Probs.10-19/20
Probs. 10-24125
10.3
10-Ui. The pump operates using the motor that has a
power of 85 W. If the impeller at Bis turning at 150 rev/min,
determine the maximum shear stress in the 20-mm-diameter
transmission shaft at A.
473
POWER T RANSMISSION
10-3L The 6-hp reducer motor can turn at 1200 rev/min.
If the allowable shear stress for the shaft is 'Tauow = 6 ksi,
determine the smallest diameter of the shaft to the nearest
1~ in. that can be used.
*10-32. The 6-hp reducer motor can turn at 1200 rev/min.
If the shaft has a diameter of in., determine the maximum
shear stress in the shaft.
i
Prob. 10-26
10-27. The gear motor can develop 1~ hp when it turns at
300 rev/min. If the shaft has a diameter of} in., determine
the maximum shear stress in the shaft.
*10-28. The gea r motor can develop 1~ hp when it turns at
80 rev/min. If the allowable shear stress for the shaft is
'Tallow= 4 ksi, determine the smallest diameter of the shaft to
the nearest k in. that ca n be used.
--
\\
-'' -"
Probs. 10-27/28
10-29. The gear motor can develop k hp when it turns at
(i()O rev/min. If the shaft has a diameter of } in.. determine
the maximum shear stress in the shaft.
10-30. The gear motor can develop 2 hp when it turns at
150 rev/min. If the allowable shear stress for the shaft is
'Tallow= 8 ksi, determine the smallest diameter of the shaft to
the nearest k in. that can be used.
Probs. 10-31132
10-33. The solid steel shaft DF bas a diameter of 25 mm
and is supported by smooth bearings al D and £. It is coupled
to a motor at F, which delivers 12 kW of power to the shaft
while it is turning at 50 rev/s. II gears A, B. and C remove
3 kW, 4 kW, and 5 kW respectively. determine the maximum
shear stress developed in the shaft within regions CF and BC.
The shaft is free to tum in its support bearings D and £.
10-34. The solid steel shaft DF has a diameter of 25 mm
and is supported by smooth bearings at D and £. It is
coupled to a motor at F, which de livers 12 kW of power to
the shaft while it is turning at 50 rcv/s. lI gears A, B, and C
remove 3 kW, 4 kW, and 5 kW respectively, determine the
absolute maximum shea r stress in the shaft.
Sk W
12kW
Lli;:~3=1k~W::;::;44k~W=~~f
2~5n:.JCll
-"Jl
IA 18 ~
D
Probs. 10-29/30
C
E
Probs. 10-33/34
F_..___.._
474
CHAPTER
10
TORSION
10.4 ANGLE OF TWIST
Long shafts subjected to torsion can, in
some cases, have a noticeable elastic
twist.
In this section we will develop a formula for determining the angle of twist
<f> (phi) of one end of a shaft with respect to its other end. To generalize this
development, we will assume the shaft has a circular cross section that can
gradually vary along its length, Fig. 10-12a. Also, the material is assumed to
be homogeneous and to behave in a linear elastic manner when the torque
is applied. As in the case of an axially loaded bar, we will neglect the localized
deformations that occur at points of application of the torques and where the
cross section changes abruptly. By Saint-Venant's principle, these effects
occur within small regions of the shaft's length, and generally they will have
only a slight effect on the final result.
Using the method of sections, a differential disk of thickness dx, located at
position x , is isolated from the shaft, Fig. 10-12b. At this location, the internal
torque is T(x), since the external loading may cause it to change along the
shaft. Due to T(x), the disk will twist, such that the relt1Live rotation of one of
its faces with respect to the other face is d<f>. As a result an element of
material located at an arbitrary radius p within the disk will undergo a shear
strain y. The values of y and d</> are related by Eq. 10-1, namely,
d<f>
=
dx
y-
(10-13)
p
z
(a)
(b)
Fig.10-U
10.4
ANGLE OF TWIST
475
Since Hooke's law, y = r / G , applies and the shear stress can be
expressed in terms of the applied torque using the torsion formula
T = T(x)p /l(x) , then
y = T(x)p/ l(x)G(x). Substituting this into
Eq. 1~13, the angle of twist for the disk is therefore
T (x)
d</> = J(x)G(x) dx
Integrating over the entire length L of the shaft, we can obtain the angle
of twist for the entire shaft, namely,
1
L
</> -
0
Here
T(x) dx
J(x)G(x)
(1~14)
= the
angle of twist of one end of the shaft with respect to the
other end, measured in radians
T(x) = the internal torque at the arbitrary position x, found from the
method of sections and the equation of moment equilibrium
applied about the shaft's axis
J(x) = the shaft's polar moment of inertia expressed as a function of x
G(x) = the shear modulus of elasticity for the material expressed as a
function of x
</>
Constant Torque and Cross-Sectional Area. Usually in
engineering practice the material is homogeneous so that G is constant.
Also, the cross-sectional area and the external torque are constant along
the length of the shaft, Fig. 1~13. When this is the case, the internal
torque T(x) = T, the polar moment of inertia J (x) = J, and Eq.1~14
can be integrated, which gives
BJ
(1~15)
Note the similarities between the above two equations and those for an
axially loaded bar.
Fig. 10-13
When calculating both the stress and
the angle of twist of this soil auger,
it is necessary to consider the
variable torsional loading which acts
along its length.
476
CHAPTER
10
TORSION
Load
dial
r-
Load
range
selector
•
Torque
strain
recorder
( Turning
head
Specimen
Motor
r:-1
........
11111
IUH
Fixed
head
Movable unit
on rails
Fig. 10-14
Equation 10-15 is often used to determine the shear modulus of
elasticity, G, of a material. To do so, a specimen of known length and
diameter is placed in a torsion testing machine like the one shown in
Fig. 10- 14. The applied torque T and angle of twist <f> are then measured
along the length L. From Eq. 10- 15, we get G = TL/l<f>. To obtain a
more reliable value of G, several of these tests are performed and the
average value is used.
Multiple Torques.
If the shaft is subjected to several different
torques, or the cross-sectional area or shear modulus changes abruptly
from one region of the shaft to the next, as in Fig. 10- 12, then Eq. 10-15
should be applied to each segment of the shaft where these quantities
are all constant. The angle of twist of one end of the shaft with respect to
the other is found from the algebraic addition of the angles of twist of
each segment. For this case,
The shear strain at points on
the cross section increases linearly
with p, i.e., y = (p/ chmax·
(b)
Fig. 10-12 (Repeated)
(10-16)
10.4
Sign Convention.
The best way to apply this equation is to use a
sign convention for both the internal torque and the angle of twist of one
end of the shaft with respect to the other end. To do this, we will apply
the right-hand rule, whereby both the torque and angle will be positive,
provided the thumb is directed outward from the shaft while the fingers
curl in the direction of the torque, Fig. 10-15.
+<f>(x)
+<f>(x)
Positive sign convention
for Tand </>
Fig.10-15
IMPORTANT POINT
• When applying Eq. 10- 14 to determine the angle of twist, it is
important that the applied torques do not cause yielding of the
material and that the material is homogeneous and behaves in
a linear elastic manner.
ANGLE OF TWIST
477
478
CHAPTER
10
TORSION
PROCEDURE FOR ANALYSIS
The angle of twist of one end of a shaft or tube with respect to the
other end can be determined using the following procedure.
Internal Torque.
• The internal torque is found at a point on the axis of the shaft
by using the method of sections and the equation of moment
equilibrium, applied along the shaft's axis.
• If the torque varies along the shaft's length, a section should be
made at the arbitrary position x along the shaft and the internal
torque represented as a function of x , i.e., T(x).
• If several constant external torques act on the shaft between
its ends, the internal torque in each segment of the shaft,
between any two external torques, must be determined.
Angle of Twist.
• When the circular cross-sectional area of the shaft varies along
the shaft's axis, the polar moment of inertia must be expressed
as a function of its position x along the axis,J(x).
• If the polar moment of inertia or the internal torque
suddenly changes between the ends of the shaft, then
</> =
T(x)/J(x)G(x)) dx or</> = TL/JG must be applied to
each segment for which J, G, and Tare continuous or constant.
J(
• When the internal torque in each segment is determined, be
sure to use a consistent sign convention for the shaft or its
segments, such as the one shown in Fig. 10-15. Also make sure
that a consistent set of units is used when substituting numerical
data into the equations.
10.4
EXAMPLE
10.5
5m
Determine the angle of twist of the end A of theA-36 steel shaft shown in
Fig. 10-16a. Also, what is the angle of twist of A relative to C? The shaft
has a diameter of200 mm.
c
SOLUTION
A
Internal Torque. Using the method of sections, the in rental rorques are
found in each segment as shown in Fig. 10-16b. By the right-hand rule,
with positive torques directed away from the sectumed end of the shaft,
we have TAB= +80 kN · m, Tse= -70 kN · m, and TcD =- 10 kN · m.
These results are also shown on the torque diagram, which indicates how
the internal torque varies along the axis of the shaft, Fig. 10-16c.
J
= 7T (0.1 m) 4 = 0.1571(10- 3 )
2
+
= 2. JG =
80kN·m
~:SOkN·m
·~:~
m4
80 kN ·m
80( 1Q3) N · m (3 m)
(0. 1571(10-3) m4 )(75(109) N/ m2)
(b)
-70(1Q3) N · m (2 m)
4
9
2
(0.1571(10- 3) m )(75( 10 ) N/ m
+
)
- 10(1Q3) N · m (1.5 m)
~~~~~~~~~~~
(0.1571(10- 3) m4)(75(109) N/rrf-)
Ans.
The re lative angle of twist of A with respect to C involves only two
segments of the shaft.
TL
(a)
Y-2~.m
For A-36 steel, the table on the back cover gives G =75 GPa. Therefore,
the end A of the shaft has a rotation of
TL
IO kN·m
60kN·m
150kN·m
80kN·m
Angle of Twist. The polar moment of inertia for the shaft is
<f>A
479
ANGLE OF TWIST
T(kN·m)
801 - - - 3
-70
80(1Q3) N · m (3 m)
__,
5
..__
I
(c)
1
cf>A/c= JG = (0.157 1(10- 3) m4)(75{109) N/ m2)
Fig. 10-16
+
-70(1Q3) N · m (2 m)
(0.1571(10- 3) m4)(75(109) N/m2 )
Ans.
Both results are positive, which means that end A will rotate as
indicated by the curl of the right-hand fingers when the thumb is
directed away from the shaft.
6.5 x(m)
- to
480
CHAPTER
EXAMPLE
10
TORSION
10.6
-
Ef
N·m ~
40
D ;~", /
Z!~, N·m ~}~ / · 5 m
150N·m B
P.~
21-:<
V
lOOmm
3
The gears attached to the fixed-end steel shaft are subjected to the torques
shown in Fig. 10-17a. If the shaft has a diameter of 14 mm, determine the
displacement of the tooth Pon gear A. G =80 GPa.
SOLUTION
;><
"'x"'o. m
/
o.4 m
(a)
By inspection, the torques in segments AC, CD, and
DE are clifferent yet constant t h roughout each segment. F ree-bo dy
diagrams of these segments along with the calculated internal torques
are shown in Fig. 10-17b. Using the right-hand rule and the established
sign convention that positive torque is directed away from the sectioned
end of the shaft, we have
Internal Torque.
TAc = + 150N · m
Angle of Twist.
~
>~~Nm~Nm
tf>A = 0.2121 rad
~~
(c)
Fig. 10-17
1T (0.007
2
m) 4
=
3.771 (10-9) m4
Applying Eq.10-16 to each segment and adding the results algebraically,
we have
TL
<f>A
=
280N·m
(b)
Tve = -170N·m
The polar moment of inertia for the shaft is
J =
Toe = 170N·m
Tcv = -130N·m
+
L JG
( + 150 N · m) (0.4 m)
=
3.771(10-9 )m4 (80(109 )N/m2]
(-130N ·m )(0.3m)
3.771 ( 10- 9 ) m4 (80 ( 109 ) N /m2]
<!>A
=
(-170N ·m )(0.5m)
+ ---~----~-~-
3.771 ( 10-9) m4 (80 ( 109 ) N /m2]
-0.2121 rad
Since the answer is negative, by the right-hand rule the thumb is directed
toward the support E of the shaft, and therefore gear A will rotate as
shown in Fig. 10-17c.
The displacement of tooth Pon gear A is
Sp
= </>Ar =
( 0.2121
rad) ( 100 mm)
=
21.2 mm
Ans.
10.4
I
EXAMPLE
481
ANGLE OF TWIST
10.7 1
The 2-in.-diameter solid cast-iron post shown in Fig.10-18a is buried 24 in.
in soil. If a torque is applied to its top using a rigid wrench, detel!111ine the
maximum shear stress in the post and the angle of twist of the wrench.
Assume that the torque is about to tum the bottom of the post, and the soil
exerts a uniform torsional resistance oft lb· in.fin. along its 24-in. buried
length. G = 5.5 ( 103 ) ksi.
"-2 ·ltl.
SOLUTION
B
Internal Torque. The internal torque in segment AB of the post is
constant. From the free-body diagram, Fig. 10- 18b, we have
IMz = O·,
TAB
=
25 lb ( 12 in.)
=
300 lb . in.
The magnitude of the uniform distribution of torque along the buried segment
BC can be determined from equilibrium of the entire post, Fig. 10-1&. Here
IMz
=
0
25 lb ( 12 in. ) - t ( 24 in.)
=
0
t = 12.5 lb· in.fin.
Hence, from a free-body diagram of the bottom segment of the post,
located at the position x, Fig. 10-18d, we have
IMz = O·,
TBc - 12.5x = 0
TBc = 12.5x
Maximum Shear Stress. The largest shear stress occurs in region AB,
since the torque is largest there and J is constant for the post. Applying
the torsion formula , we have
25lb
TAB c
( 300 lb . in.) ( 1 in.)
.
Ans.
'Tmax =
J =
) ( . )4
= 191 psi
( 1T f2
1 m.
Angle of Twist. The angle of twist at the top can be determined
relative to the bottom of the post, since it is fixed and yet is about to tum.
Both segments AB and BC twist, and so in this case we have
<f>A
1
TAB LAB +
= ---
JG
Lsc
0
TBc dx
JG
_ (300lb ·in. )36in. + (
JG
Jo
=
I
36 in.
24
0
i
12.5xdx
JG
10 800 lb . in2
12.5( ( 24) 2f2] lb . in2
JG
+
JG
- @
(w T se
l (w
x
(~ r = 12.5 lb·in./in.
14 400 lb. in2
( 1Tf2) (1 in.) 4 5500(1a3) lbfin2
=
0.00167 rad
Ans.
(d)
Fig.10-18
482
CHAPTER
10
TORSION
FUNDAMENTAL PROBLEMS
Fl0-9. The 60-mm-diameter steel shaft is subjected to the
torques shown. Determine the angle of twist of end A with
respect to C. Take G = 75 GPa.
Fl0-12. A series of gears are mounted on the40-mm-diameter
steel shaft. D etermine the angle of twist of gear E relative
to gear A. Take G = 75 GPa.
2kN·m
Prob.Fl0-9
Fl0-10. Determine the angle of twist of wheel B with
respect to wheel A. The shaft has a diameter of 40 mm and
is made of steel for which G = 75 GPa.
Prob. Fl0-12
Fl0-13. The SO-mm-diameter shaft is made of steel. If it is
subjected to the uniform distributed torque, determine the
angle of twist of end A. Take G = 75 GPa.
B
4 kN lOkN
2kN
Prob.Fl0-10
Fl0-11. The hollow 6061-T6 aluminum shaft has an outer
and inner radius of c,, = 40 mm and C; = 30 mm, respectively.
Determine the angle of twist of end A. The support at Bis
flexible like a torsional spring, so that Ts = ks <f>s, where the
torsional stiffness is ks = 90 kN · m/rad.
A
Prob. Fl0-13
Fl0-14. Tue SO-mm-diameter shaft is made of steel. If it is
subjected to the triangular distributed load, determine the
angle of twist of end A. Take G = 75 GPa.
A 3kN·m
Prob.Fl0-11
Prob. Fl0-14
10.4
483
ANGLE OF TWIST
PROBLEMS
10-35. The propellers of a ship arc connected to an A-36
steel shaft that is 60 m long and has an outer diameter of
340 mm and inner diameter of 260 mm. If the power output
is 4.5 MW when the shaft rotates at 20 rad/s, determine the
maximum torsional stress in the shaft and its angle of twist.
10-38. The A-36 steel shaft has a diameter of 50 mm and is
subjected to the distributed and concentrated loadings
shown. Determine the absolute maximum shear stress in
the shaft and plot a graph of the angle of twist of the shaft in
radians versus x.
*10-36. The solid shaft of radius c is subjected to a torque T
at its ends. Show that the maximum shear strain in the shaft is
'Ymax = Tc/JG. What is the shear strain on an clement located
at point A, c/2 from the center of the shaft? Sketch the shear
strain distortion of this clement.
c/2
T
Prob. 10-38
Prob. 10-36
10-37. The splined ends and gears attached to the A992
steel shaft arc subjected to the torques shown. Determine
the angle of twist of end 8 with respect to end A. The shaft
has a diameter of 40 mm.
10-39. The 60-mm-diamctcr shaft is made of 6061-T6
aluminum having an allowable shear stress of 'Tallow = 80 MPa.
Determine the maximum allowable torque T. Also, find the
corresponding angle of twist of disk A relative to disk C.
*10-40. The 60-mm-diametcr shaft is made of 6061-T6
aluminum. If the allowable shear stress is 'Tn11ow = 80 MPa,
and the angle of twist of disk A relative to disk C is limited
so that it does not exceed 0.06 rad, determine the maximum
allowable torque T.
400N·m
B
A
200 N·m
500 mm
//
400mm
IT
3
Prob. 10-37
Probs. 10-39/40
484
CHAPTER
10
TORSION
10--41. The SO-mm-diameter A992 steel shaft is subjected
to the torques shown. Determine the angle of twist of the
end A.
*10--44. The rotating flywheel-and-shaft, when brought to
a sudden stop at D. begins to oscillate clockwise-counterclockwise such that a point A on the outer edge of the
fly-wheel is displaced through a 6-mm arc. Determine the
maximum shear stress developed in the tubular A-36 steel
shaft due to this oscillation. The shaft has an inner diameter
of 24 mm and an outer diameter of 32 mm. The bearings at
Band Callow th e shaft to rotate freely, whereas the support
at D holds the shaft fixed.
Prob. 10--41
Prob. 10--44
10--42. The shaft is made of A992 steel with the allowable
shear stress of Ta11ow = 75 MPa. If gear B supplies 15 kW of
power. while gears A , C and D withdraw 6 kW. 4 kW and
5 k\V. respectively, determine the required minimum
diameter d of the shaft to the nearest millimeter. Also. find
the corresponding angle of twist of gear A relative to gear
D. The shaft is rotating at 600 rpm.
10--43. Gear B supplies 15 kW of power. while gears A , C,
and D withdraw 6 kW, 4 kW and 5 kW, respectively. If the
shaft is made of steel with the allowable shear stress of
Tallow = 75 MPa, and the relative angle of twist between any
two gears cannot exceed 0.05 rad, determine the required
minimum diameter d of the shaft to the nea rest millimeter.
The shaft is rotating at 600 rpm.
Probs. 10--42/43
10--45. The turbine develops 150 kW of power, which is
transmitted to the gears such that C receives 70% and D
receives 30%. If the rotation of the 100-mm-diameter A-36
steel shaft is "' = 800 rev / min., determine the absolute
maximum shear stress in the shaft and the angle of twist of
end E of the shaft relative to B. The journal bearing at E
allows the shaft to turn freely about its axis.
10--46. The turbine develops 150 kW of power, which is
transmitted to the gears such that both C and D receive an
equal amount. 1f the rotation of the 100-mm-diameter A-36
steel shaft is "' = 500 rev/min., determine the absolute
maximum shear stress in the shaft and the rotation of end B
of the shaft relative to £.The journal bearing at E allows
the shaft to turn freely about its axis.
Probs. 10--45146
10.4
10-47. The shaft is made of A992 steel. It has a diameter
of 1 in. and is supported by bearings at A and D , which
allow free rotation. Determine the angle of twist of B with
respect to D.
*10-48. The shaft is made of A-36 steel. It has a diameter
of 1 in. and is supported by bearings at A and D , which
allow free rotation. Determine the angle of twist of gear C
with respect to 8.
Probs. 10-47/48
10-49. The A992 steel shaft has a diameter of 50 mm and
is subjected to the distributed loadings shown. Determine
the absolute maximum shear stress in the shaft and plot a
graph of the angle of twist of the shaft in radians versus x.
ANGLE OF TWIST
10-50. The turbine develops 300 kW of power, which is
transmitted to the gears such that both 8 and C receive an
equal amount. If the rotation of the 100-mm-diameter A992
steel shaft is w = 600 rev/min., determine the absolute
maximum shear stress in the shaft and the rotation of end D
of the shaft relative to A. The journal bearing at D allows
the shaft to tum freely about its axis.
Prob. 10-50
10-SL The device shown is used to mix soils in order to
provide in-situ stabilization. If the mixer is connected to an
A-36 st eel tubular shaft that has an inner diameter of 3 in.
and an outer diameter of 4.5 in., determine the angle of
twist of the shaft at A relative to C if each mixing blade is
subjected to the torques shown.
15 ft
5000 lb· ft
A
~
Prob. 10-49
485
Prob. 10-51
1
486
CHAPTER
10
TORSION
*10-52. The device shown is used to mix soils in order to
provide in-situ stabilization. If the mixer is connected to an
A-36 steel tubular shaft that has an inner diameter of 3 in.
and an outer diameter of 4.5 in, determine the angle of twist
of the shaft at A relative to B and the absolute maximum
shear stress in the shaft if each mixing blade is subjected to
the torques shown.
10-54. The A-36 hollow steel shaft is 2 m Jong and has an
outer diameter of 40 mm. When it is rotating at 80 rad/s, it
transmits 32 kW of power from the engine E to the
generator G. Determine the smallest thickness of the shaft
if the allowable shear stress is r allow = 140 MPa and the shaft
is restricted not to twist more than 0.05 rad.
10-55. The A-36 solid steel shaft is 3 m Jong and has a
diameter of 50 mm. It is required to transmit 35 kW of
power from the engine E to the generator G. Determine the
smallest angular velocity of the shaft if it is restricted not to
twist more than 1°.
Probs. 10-54155
Prob.10-52
10-53. The 6-in.-diameter L-2 steel shaft on the turbine is
supported on journal bearings at A and B. If C is held fixed
and the turbine blades create a torque on the shaft that
increases linearly from zero at C to 2000 lb · ft at D ,
determine the angle of twist of the shaft at D relative to C.
Also, calculate the absolute maximum shear stress in the
shaft. Neglect the size of the blades.
*10-56. The shaft of radius c is subjected to a distributed
torque 1, measured as torque/length of shaft. Determine the
angle of twist at end A. The shear modulus is G.
2 ft
B
Prob.10-53
Prob.10-56
10.4
10-57. The A-36 steel bolt is tightened within a hole so
that the reactive torque on the shank AB can be expressed
by the equation t = (kx2) N · m/m, where xis in meters. If a
torque of T = 50 N · m is applied to the bolt head,
determine the constant k and the amount of twist in the
50-mm length of the shank. Assume the shank has a constant
radius of 4 mm.
ANGLE OF TWIST
4 87
The 60-mm diameter solid shaft is made of2014-T6
aluminum and is subjected to the distributed and
concentrated torsional loadings shown. Determine the angle
of twist at the free cod A of the shaft.
*10-60.
10-58. Solve Prob. 10-57 if the distributed torque is
2
t = (kx /3) N · m/m.
/B
0.4m~
0.6111~
Prob. 10-60
Probs. 10-57/58
10-59. The tubular drive shaft for the propeller of a
hovercraft is 6 m long. If the motor delivers 4 MW of power
to the shaft when the propellers rotate at 25 rad/s, determine
the required inner diameter of the shaft if the outer
diameter is 250 mm. What is the angle of twist of the shaft
when it is operatiog?Take 'Tat1ow= 90 MPa and G = 75 GPa.
10-61. The motor produces a torque of T = 20 N • m on
gear A. If gear C is suddenly locked so it does not turn, yet B
can freely turn. determine the angle of twist of F with respect
to E and F with respect to D of the L2-steel shaft, which bas
an inner diameter of 30 mm and an outer diameter of 50 mm.
Also. calculate the absolute maximum shear stress in the
shaft. The shaft is supported on journal bearings at G at H.
-6m
IOOmm
D
~
Prob. 10-59
c
·1- - - 0.8 m - - -1-0.4 m - m
02m
Prob. 10-61
488
CHAPTER
10
TORSION
10.5
STATICALLY INDETERMINATE
TORQUE-LOADED MEMBERS
A torsionally loaded shaft will be statically indeterminate if the moment
equation of equilibrium, applied about the axis of the shaft, is not adequate
to determine the unknown torques acting on the shaft. An example of this
situation is shown in Fig. 10- 19a. As shown on the free-body diagram,
Fig. 10-19b, the reactive torques at the supports A and Bare unknown.
Along the axis of the shaft, we require
2-M = O·,
In order to obtain a solution, we will use the same method of analysis
discussed in Sec. 9.4. The necessary compatibility condition requires the
angle of twist of one end of the shaft with respect to the other end to be
equal to zero, since the end supports are fixed. Therefore,
<f>AJB = 0
Provided the material is linear elastic, we can then apply the
load-displacement relation</> = TL/JG to express this equation in terms
of the unknown torques. Realizing that the internal torque in segment AC
is + 1A and in segment CB it is -TB, Fig. 10-19c, we have
TA(3 m) _ TB(2 m)
JG
JG
=
0
Solving the above two equations for the reactions, we get
TA
=
200 N · m
and
(c)
Fig.10-19
TB
=
300 N · m
10.5
STATICALLY INDETERMINATE TORQUE-LOADED M EMBERS
PROCEDURE FOR ANALYSIS
The unknown torques in statically indeterminate shafts are
dete rmined by satisfying equilibrium, compatibility, and
load-displacement requirements for the shaft.
Equilibrium.
• Draw a free-body diagram of the shaft in order to identify all
the external torques that act on it. Then write the equation of
moment equilibrium about the axis of the shaft.
Compatibility.
• Write the compatibility equation. Give consideration as to how
the supports constrain the shaft when it is twisted.
Load-Displacement.
• Express the angles of twist in the compatibility condition in
terms of the torques, using a load-displacement relation, such
as</> = TL/JG.
• Solve the equations for the unknown reactive torques. If any of
the magnitudes have a negative numerical value, it indicates
that this torque acts in the opposite sense of direction to that
shown on the free-body diagram.
The shafl of this cutting machine is Cixed at its
ends and subj ected to a torque at its center,
allowing it to act as a torsiona l spring.
489
490
CHAPTER
EXAMPLE
10
TORSION
10.8
-
-
The solid steel shaft shown in Fig. 10--20a has a diameter of 20 mm. If it is
subjected to the two torques, determine the reactions at the fixed supports
A andB.
A
B
(a)
(b)
SOLUTION
Equilibrium. By inspection of the free-body diagram, Fig. l0--20b, it is
seen that the problem is statically indeterminate, since there is only one
available equation of equilibrium and there are two unknowns. We require
IM, = 0;
-Ta+800N · m-500N · m-TA = O
(1)
Compatibility. Since the ends of the shaft are fixed, the angle of twist
of one end of the shaft with respect to the other must be zero. Hence, the
compatibility equation becomes
800 - Ts
</>A/a
300 - Ts
(c)
Fig. 10-20
= 0
Load-Displacement. This condition can be expressed in terms of the
unknown torques by using the load-displacement relationship, 4> = TL/JG.
Here there are three regions of the shaft where the internal torque is
constant. On the free-body diagrams in Fig. 10--20c we have shown the
internal torques acting on the left segments of the shaft. This way the internal
torque is only a function of Ta. Using the sign convention established in
Sec. 10.4, we have
-Ta(0.2 m) + ( 800 - Ta) ( 1.5 m) + (300 - Ta)( 0.3 m)
JG
JG
JG
so that
Ta = 645 N · m
=
0
Ans.
Using Eq.1,
TA = -345 N · m
Ans.
The negative sign indicates that T A acts in the opposite direction of that
shown in Fig. l0- 20b.
10.5
EXAMPLE
491
STATICALLY INDETERMINATE TORQUE-LOADED M EMBERS
10.9
The shaft shown in Fig. L0-2 la is made from a steel tube, which is bonded
to a brass core. lf a torque of T = 250 lb · ft is applied at its end, plot the
shear stress distribution alo ng a radial line on its cross section. Take
G,t = 11.4 ( H>3) ks i, Gbr = 5.20 ( H>3) ksi.
8
SOLUTION
Equilibrium. A free-body diagram of the shaft is shown in Fig. 10-21b. ( ,
The reactio n a t the wa ll has been re presented by the amount of torque
inA
·
resisted by the steel, T.t, and by the brass, Tb,. Working in units of pounds r- 250 lb·fi1
and inches, e quilibrium requires
-T.t - Tbr + ( 2501b·ft )( 12in./ ft) = 0
(1)
\I
(a)
Compatibility. We require the angle of twist of end A to be the same
for both the steel and brass since they are bonded together. Thus,
</J
Load-Displacement.
</J = TL/JG,
( 7r/ 2)[ ( 1 in. ) 4
-
= c/Jst = c/Jbr
Applying the load- displacement relationship,
250 lb· rt
( 0.5 in. ) 4} ll.4(1a3) kip/ in2
T.t
(b)
Tb,L
( 7r/ 2 ) ( 0.5 in. ) 4 5.20 ( 10 3 ) kip/ in2
~~~~~~..:..:..._~~~~~~
= 32.88 Tbr
(2)
Solving Eqs. 1 and 2, we get
T.t = 2911.5 lb · in. = 242.6 lb · ft
Tbr = 88.5 lb· in. = 7.38 lb · ft
The shear stress in the brass core varies from zero at its cente r to a
maximum at the inte rface where it contacts the steel tube. Using the
torsio n fo rmula,
( 88.5 lb · in. ) ( 0.5 in. )
.
=
=
451 psi
( 7i )
br max
( 7T12 ) ( 0.5 in. ) 4
.
For the steel, the minimum and maximum shear stresses are
.
(29 11.5 lb· in. ) ( 0.5 in. )
= 989 psi
( -r. ) · =
st mm
( 7r/2 )[ ( 1 in. ) 4 - ( 0.5in.)4}
(
-r.)
st
=
max
1977 psi
0.5 in.
Shcar-5trcss distribution
(c)
( 29ll.5lb·in.){lin.)
= l 9??psi
( 7T / 2 )[ ( 1 in.) 4 - ( 0.5 in.) 4)
The results are plotted in Fig. 10-21c. Note the discontinuity of shear
stress at the brass and steel interface. This is to be expected, since the
materials have differe nt moduli of rigidity; i.e., steel is stiffer than brass
( Gst > Gbr), and thus it carries more shear stress at the interface.
Although the shear stress is discontinuous here, the shear strain is not.
Rather, it is the same o n e ither side of the brass-steel interface, Fig.10-21d.
Shcar-5train distribution
(d)
Fig. 10-21
492
CHAPTER
10
TORSION
PROBLEMS
10-62. The steel shaft has a diameter of 40 mm and is fixed
at its ends A and B. If it is subjected to the couple, determine
the maximum shear stress in regions AC and CB of the
shaft. Gsi = 75 GPa.
3kN
A
10-65. The bronze C86100 pipe has an outer diameter of
1.5 in. and a thickness of 0.125 in. The coupling on it at C is
being tightened using a wrench. If the torque developed at
A is 125 lb · in., determine the magnitude F of the couple
forces. The pipe is fixed supported at end B.
10-66. The bronze C86100 pipe has an outer diameter of
1.5 in. and a thickness of 0.125 in. The coupling on it at C is
being tightened using a wrench. If the applied force is
F = 20 lb, determine the maximum shear stress in the pipe.
B
Prob. 10-62
10-63. The A992 steel shaft has a diameter of 60 mm and is
fixed at its ends A and B. If it is subjected to the torques shown,
determine the absolute maximum shear stress in the shaft.
B
Probs. 10-65/66
Prob. 10-63
*10-64. The steel shaft is made from two segments: AC
has a diameter of 0.5 in., and CB has a diameter of 1 in. If
the shaft is fixed at its ends A and B and subjected to a
torque of 500 lb · ft, determine the maximum shear stress
in the shaft. Gsi = 10.8(103) ksi.
A
10-67. The shaft is made of L2 tool steel, has a diameter of
40 mm, and is fixed at its ends A and B. If it is subjected to
the torque, determine the maximum shear stress in regions
AC and CB.
0.5 in.
A
Prob. 10-64
Prob. 10-67
10.5
STATICAUY INDETERMINATE TORQUE-LOADED MEMBERS
*10-68. The shaft is made of L2 tool steel, has a diameter
of 40 mm. and is fixed at its ends A and B. If it is subjected
to the couple. determine the maximum shear stress in
regions AC and CB.
493
10-71. The two shafts are made of A-36 steel. Each has a
diameter of 25 mm and they are connected using the gears
fixed to their ends. Their other ends arc attached to fixed
supports at A and 8. They are also supported by journal
bearings at C and D, which allow Cree rotation of the shafts
along their axes. If a torque of 500 N · m is applied to the
gear at £,determine the reactions at A and B.
2kN
10-72. The two shafts are made of A-36 steel. Each has a
diameter of 25 mm and they are connected using the gears
fixed to their ends. Their other ends arc attached to fixed
supports at A and 8. They are also supported by journal
bearings at C and D , which allow free rotation of the shafts
along their axes. If a torque of 500 N ·mis applied to the gear
at £,determine the rotation of this gear.
Prob.10-68
10-69. TheAm I004-T61 magncsium tube is bonded to the
A-36 steel rod. If the allowable shear stresses for the
magnesium and steel are (TaJlow)mg = 45 MPa and (Tanow)s1 =
75 MPa, respectively, determine the maxin1um allowable
torque that can be applied at A.Also,find the corresponding
angle of twist of end A.
10-70. The Aml004-T61 magnesium tube is bonded to the
A-36 steel rod. If a torque of T = 5 kN ·mis applied to end A ,
determine the maximum shear stress in each material.
Sketch the shear stress distribution.
Probs. 10-71n2
10-73. A rod is made from two segments: AB is steel and
BC is brass. It is fixed at its ends and subjected to a torque of
T = 680 N · m. II the steel portion has a diameter of 30 mm,
determine the required diameter of the brass portion so the
reactions at the walls will be the same. G51 = 75 GPa,
Gbr = 39GPa.
10-74. D etermine the absolute maxinlum shear stress in
the shaft of Prob. 10--73.
c
Probs. 10-69no
Probs. 10-73n4
494
10
CHAPTER
TORSION
10-75. The two 3-ft-long shafts are made of 2014-16
aluminum. Each has a diameter of 1.5 in. and they are
connected using the gears fixed to their ends. Their other
ends are attached to fixed supports at A and B. They are also
supported by bearings at C and D , which allow free rotation
of the shafts along their axes. If a torque of 600 lb· ft is
applied to the top gear as shown, determine the maximum
shear stress in each shaft.
10-77. If the shaft is subjected to a uniform distributed
torque of 1 = 20 kN · m/m, determine the maximum shear
stress developed in the shaft. The shaft is made of 2014-T6
aluminum alloy and is fixed at A and C.
'--..._
400mm
'--..._
~
600 mm
~"-'-.
a
8~1
A
3 ft
~
60mm ~
/
Section a-a
Prob.10-77
10-78. The tapered shaft is confined by the fixed supports
at A and B. If a torque T is applied at its mid-point,
determine the reactions at the supports.
2 in.
Prob.10-75
T
A
*10-76. The composite shaft consists of a mid-section that
includes the 1-in.-diameter solid shaft and a tube that is
welded to the rigid flanges at A and B. Neglect the thickness
of the flanges and determine the angle of twist of end C of
the shaft relative to end D. The shaft is subjected to a torque
of 800 lb · ft. The material is A-36 steel.
L/2
l
c
L/
Prob.10-78
10-79. The shaft of radius c is subjected to a distributed
torque 1, measured as torque/length of shaft. Determine the
reactions at the fixed supports A and B.
B
800~='=1;,;:n.:{-.:Jt_.::3:J:n.===0.=25t/:-;in.
~-
-o.s
800 lb· ft
ft:1-=:.~~~~)Ir--.--t f11v
A
-o.Sft
Prob.10-76
Prob.10-79
CHAPTER REVIEW
CHAPTER REVIEW
Torque causes a shaft having a circular cross
section to twist, such that whatever the torque,
the shear strain in the shaft is always proportional
to its radial distance from the center of the shaft.
Provided the material is homogeneous and
linear elastic, then the shear stress is
determined from the torsion formula,
'T
= -
Tp
J
The design of a shaft requires finding the
geometric parameter,
J
T
C
'Tallow
Often the power P supplied to a shaft rotating
at w is reported, in which case the torque is
determined from P = Tw.
The angle of twist of a circular shaft is
determined from
</>
=
LT(x) dx
1
0
J(x)G(x)
If the internal torque and JG are constant
within each segment of the shaft then
For application, it is necessary to use a sign
convention for the internal torque and to be
sure the material remains linear elastic.
If the shaft is statically indeterminate, then the
reactive torques are determined from
equilibrium, compatibility of twist, and a
load-displacement relationship, such as
</>=TL/JG.
T
495
496
10
CHAPT ER
TORSION
REVIEW PROBLEMS
RlO-L The shaft is made of A 992 steel and has an allowable
shear stress of r allow = 75 MPa. when the shaft is rotating at
300 rpm, the motor supplies 8 kW of power, while gears A
and B withdraw 5 kW and 3 kW, respectively. Determine the
required minimum diameter of the shaft to the nearest
millimeter. Also, find the rotation of gear A relative to C.
*Rl0-4. The shaft has a radius c and is subjected to a torque
per unit length of 10, which is distributed uniformly over the
shaft's entire length L. If it is fixed at its far end A , determine
the angle of twist <f> of end B. The shear modulus is G.
Rl0-2. The shaft is made of A992 steel and has an
allowable shear stress of r allow = 75 MPa. when the shaft is
rotating at 300 rpm, the motor supplies 8 kW of power,
while gears A and B withdraw 5 kW and 3 kW, respectively.
If the angle of twist of gear A relative to C is not allowed to
exceed 0.03 rad, determine the required minimum diameter
of the shaft to the nearest millimeter.
A
~
300mm
Prob. Rl0-4
300mm
~
Probs. Rl0-112
Rl0-5. The motor delivers 50 hp while turning at a
constant rate of 1350 rpm at A. Using the belt and pulley
system this Loading is delivered to the steel blower shaft BC.
Determine t o the nearest ! in. the smallest diameter of this
shaft if the allowable shear stress for steel is r allow = 12 ksi.
Rl0-3. The A-36 steel circular tube is subjected to a torque
of 10 kN · m. Determine the shear stress at the mean radius
p = 60 mm and calculate the angle of twist of the tube if it is
4 m long and fixed at its far end. Solve the problem using
Eqs. 10-7 and 10-15 and using Eqs. 10-18 and 10-20.
c
p = 60mm
'\
(
A
2inT
Prob. Rl0-3
Prob. Rl0-5
REVIEW PROBLEMS
Rl~.
Segments AB and BC of the assembly are made
from 6061-T6 aluminum and A992 steel, respectively. U
couple forces P = 3 kip are applied to the lever arm,
determine the maximum shear stress developed in each
segment. The assembly is fixed at A and C.
497
*Rl0-8. The tapered shaft is made from 2014-T6
aluminl!lm alloy, and has a radius which can be described by
the equation r = 0.02(1 + x3'2) m, where x is in meters.
Determine the angle of twist of its end A if it is subjected to
a torque of 450 N · m.
r = 0.02( I + x3f2) m
450 N·m
4 in.
p
Prob.
Prob. Rl0-8
Rl~
Rl0-7. Segments AB and BC of the assembly are made
from 6061-T6 aluminum and A992 steel, respectively.Uthe
allowable shear stress for the aluminum is (ra11o,.,)a1 = 12 ksi
and for the steel (r811.,,.),. = 10 ksi, determine the maximum
allowable couple forces P that can be applied to the lever
arm. The assembly is fixed at A and C.
Rl0-9. The 61l-mm-diameter shaft rotates at 300 rev /min.
This motion is caused by the unequal belt tensions on the
pulley of 800 N and 450 N. Determine the power transmitted
and the maximum shear stress developed in the shaft.
./
300 revfmin
100mm
450 N
SOON
Prob. Rl0-7
Prob. Rl0-9
CHAPTER
1 1
(© Construction Photography/Corbis)
The girders of this bridge have been designed on the basis of their ability
to resist bending stress.
BENDING
CHAPTER OBJECTIVES
•
To represent the internal shear and moment in a beam or shaft as
a function of x.
•
To use the re lations between distributed load, shear, and moment
to draw shear and moment diagrams.
•
To determine the stress in elastic symmetric members subject to
bending.
•
To develop methods to determine the stress in unsynnmetric
beams subject to bending.
11.1
SHEAR AND MOMENT DIAGRAMS
Members that are slender and support loadings that are applied
perpendicular to their longitudinal axis are called beams. In general,
beams are long, straight bars having a constant cross-sectional area. Often
they are classified as to how they are supported. For example, a simply
supported beam is pinned at one end and roller supported at the other,
Fig.11- 1, a cantilevered beam is fixed at one end and free at the other, and
an overhanging beam has one or both of its ends freely extended over the
499
500
CHAPTER
11
BENDING
• •
Simply supported beam
Cantilevered beam
,
'
Overhanging beam
Fig.11-1
A
p
~
--IVo
___.~
__
t" I .,_;J_c_,
llS
Fig.11-2
Positive external distributed load
iv
Positive internal shear
M
M
(
'
Positive internal moment
Beam sign convention
Fig.11-3
D
supports. Beams are considered among the most important of all structural
elements. They are used to support the floor of a building, the deck of a
bridge, or the wing of an aircraft. Also, the axle of an automobile, the
boom of a crane, even many of the bones of the body act as beams.
Because of the applied loadlings, beams develop an internal shear force
and bending moment that, in general, vary from point to point along the
axis of the beam. In order to properly design a beam it therefore becomes
important to determine the maximum shear and moment in the beam.
One way to do this is to express V and M as functions of their arbitrary
position x along the beam's axis, and then plot these functions. They
represent the shear and moment diagrams, respectively. The maximum
values of V and M can then be obtained directly from these graphs. Also,
since the shear and moment diagrams provide detailed information
about the variation of the shear and moment along the beam's axis, they
are often used by engineers to decide where to place reinforcement
materials within the beam or how to proportion the size of the beam at
various points along its length.
In order to formulate V and! Min terms of x we must choose the origin
and the positive direction for x. Although the choice is arbitrary, most
often the origin is located at t he left end of the beam and the positive x
direction is to the right.
Since beams can support portions of a distributed load and
concentrated forces and couple moments, the internal shear and
moment functions of x will be discontinuous, or their slopes will be
discontinuous, at points where the loads are applied. Because of this,
these functions must be determined for each region of the beam
between any two discontinuities of loading. For example, coordinates
x i> Ni, and x3 will have to be used to describe the variation of V and M
throughout the length of the beam in Fig. 11- 2. Here the coordinates are
valid only within the regions from A to B for x1, from B to C for Ni, and
from C to D for x3 .
Beam
Sign
Convention. Before presenting a method for
determining the shear and moment as functions of x , and later plotting
these functions (shear and moment diagrams), it is first necessary to
establish a sign convention in order to define "positive" and "negative"
values for V and M. Although the choice of a sign convention is arbitrary,
here we will use the one often used in engineering practice. It is shown in
Fig. 11- 3. The positive directions are as follows: the distributed load acts
upward on the beam, the internal shear force causes a clockwise rotation
of the beam segment on which it acts, and the internal moment causes
compression in the top fibers of the segment such that it bends the
segment so that it "holds water." Loadings that are opposite to these are
considered negative.
11.1
SHEAR ANO M OMENT DIAGRAMS
501
IMPORTANT POINTS
• Beams are long straight members that are subjected to loads perpendicular to their longitudinal axis. They
are classified according to the way they are supported, e.g., simply supported, cantilevered, or overhanging.
• In order to properly design a beam, it is important to know the variation of the internal shear and moment
along its axis in order to find the points where these values are a maximum.
• Using an established sign convention for positive shear and moment, the shear and moment in the beam
can be determined as a function of their position x on the beam, and then these functions can be plotted
to form the shear and moment diagrams.
PROCEDURE FOR ANALYSIS
The shear a nd moment diagrams for a beam can be constructed using the following procedure.
Support Reactions.
• Determine all the reactive forces and couple moments acting on the beam, and resolve all the forces into
components acting perpendicular and parallel to the beam's axis.
Shear and Moment Functions.
• Specify separate coordinates x having an origin at the beam's left end and extending to regions of the
beam between concentrated forces and/or couple moments, or where there is no discontinuity of
distributed loading.
• Section the beam at each distance x, and draw the free-body diagram of one of the segments. Be sure
V and M are shown acting in their positive sense, in accordance with the sign convention given in Fig.11-3.
• The shear is obtained by summing forces perpendicular to the beam's axis.
• To eliminate V, the moment is obtained directly by summing moments about the sectioned end of
the segment.
Shear and Moment Diagrams.
• Plot the shear diagram (V versus x) and the moment diagram (M versus x). If nume rical values of the
functions describing V and M are positive, the values are plotted above the x axis, whereas negative
values are plotted below the axis.
• Generally it is conve nient to show the shear and moment diagrams below the free-body diagram of
the beam.
502
I
CHAPTER
EXAMPLE
11
BE ND I NG
11.1
Draw the shear and moment diagrams for the beam shown in Fig. ll-4a.
3kN/m
ttt} ttttttttttttt
CL
• L
I
SOLUTION
•
4m - - - - I
Support Reactions.
The support reactions are shown in Fig. 11-4c.
(a)
A free-body diagram of the left segment
of the beam is shown in Fig. 11-4b. The distributed loading on this segment
is represented by its resultant force (3x) kN, which is found only after the
segment is isolated as a free-body diagram. This force acts through the
centroid of the area under the distributed loading, a distance of x/2 from
the right end. Applying the two equations of equilibrium yields
Shear and Moment Functions.
6 kN
+ jlF.y
=
6 kN - (3x) kN - V
O·'
=
0
(b)
V = (6 - 3x) kN
3kN/m
6 kN
4m
-6 kN(x) + (3x) kN (~x) + M
~+IM = O;
M
6kN
=
(6x - 1.5x2 ) kN · m
(1)
=
0
(2)
Shear and Moment Diagrams. The shear and moment diagrams
V(kN}
shown in Fig.11-4c are obtained !by plotting Eqs.1and2.The point of zero
shear can be found from Eq. 1:
6
1----...:::.....,...:::::----+-x(m)
V
-6
=
(6 - 3x) kN
=
0
x = 2m
M (kN·m)
1~ ~
(c)
Fig.11-4
NOTE: From the moment diagram, this value of x represents the point on
x (m)
the beam where the maximum moment occurs, since by Eq. 11- 2
(see Sec. 11.2) the slope V = dM / dx = 0. From Eq. 2, we have
Mmax
=
(6 (2) - 1.5 (2) 2 ] kN · m
=
6kN · m
11.1
EXAMPLE
50 3
SHEAR ANO M OMENT DIAGRAMS
11 .2
Draw the shear and mome nt diagrams for the beam shown in Fig. ll-5a.
2 kN/m
•
3 kN
(1(---------
2 kN/m
_.........-- --------::
6 kN·m l -2m
j
J
{b)
(a)
2
.l(
x)x
2 3
SOLUTION
2
____ IV
, = -3 X
Support Reactions. The distributed load is replaced by its resultant
force, and the reactio ns have bee n determined, as shown in Fig. 11-5b.
(f~------- i.,M
-~x-IV
6 kN·m
Shear and Moment Functions. A free-body diagram of a beam
segment of le ngth xis shown in Fig. 11- 5c. The intensity of the triangular
load at the sect.ion is found by proportion, that is, w/x = (2 kN / m) /3 m
or w = (sx) kN / m. The resultant of the distributed loading is found
from the area unde r the diagram. Thus,
3kN -~ (~x)x - V = O
+f IF, =O;
1--X
(c)
2kN/m
3kqpgrrrTIJ
6 kN·m' t t
v=(3-~x2)kN
C+IM
= O;
6 kN · m - (3 kN) (x) +
M
=
(-6+ 3x - ~x3 )
~ (~ x) x (! x) +
V(k N)
(1)
3t-- - - --........ x(m)
M= 0
~p
l--·m_)_ _ __=
M
kN·m
(2)
Shear and Moment Diagrams. The graphs of Eqs. 1 and 2 are shown
in Fig. 11-5d.
(d)
Fig. 11-5
,_.._;..x (m)
504
I
CHAPTER
EX AMPLE
11
BE N DI N G
11.3 1
16
£TI i i i i r J l l
kip/ft
2 kip/ ft
1 - - - -18 ft - - - - 1
(a)
36 kip 36 kip
--------: 4 kip/ ft
----
..=--=:.::.:-:..-.: -, - - - - 11I 2 kip/ft
9ft- 1
12 ft
I
18 ft
30kip
42 kip
Draw the shear and moment diagrams for the beam shown in Fig.11-6a.
SOLUTION
Support Reactions. The distributed load is divided into triangular and
rectangular component loading&, and these loadings are then replaced by
their resultant forces. The reactions have been determined as shown on
the beam's free-body diagram, Fig.11-6b.
A free-body diagram of the left
segment is shown in Fig. 11-6c. As above, the trapezoidal loading is
replaced by rectangular and triangular distributions. Here the intensity
of the triangular load at the section is found by proportion. The resultant
force and the location of eac!h distributed loading are also shown.
Applying the equilibrium equations, we have
Shear and Moment Functions.
+ jlF.y
=
O·'
30 kip - (2 kip/ft)x -
v=
~(4 kip/ft)( 1; ft)x - V =
0
2
~ ) kip
( 30 - 2x -
(1)
~+ IM = O;
6 kip/ft
-30kip(x) +
(2kip/ft)x(~) + ~(4kip/ft)(l;ft)x(~)
+M = 0
2 kip/ft
M = ( 30x - x
f
30 kip
I
v (kip)
30
42 kip
2
-
~;)kip · ft
Equations 1 and 2 are plotted in
Fig. 11-6d. Since the point of maximum moment occurs when
dM/ dx = V = 0 (Eq.11- 2), then, from Eq.1,
Shear and Moment Diagrams.
x2
V = 0 = 30-2x- 9
Choosing the positive root,
x = 9.735 ft
(d}
Fig.11-6
(2)
Thus, from Eq. 2,
(9.735) 3
= 30(9.735) - (9.735)- = 163 kip · ft
27
?
Mmax
11.1
-
EXAMPLE
11.4
-
Draw the shear and moment diagrams for the beam shown in Fig.11- 7a.
lSkN
~
tt
&OkN·m
Aeii
SOLUTION
Support Reactions. The reactions at the supports are shown on the
free-body diagram of the beam, Fig. ll- 7d.
Shear and Moment Functions. Since there is a discontinlllity of
distributed load and also a concentrated load at the beam's .center,
two regions of x must be considered in order to describe the shear
and moment functions for the entire beam.
0
~ X1
Bi
S.7S kN
(b)
lS kN S kN/m (x2
80kN·m
v=
5.75 kN -
0
" - -5m ---i---i---i v
(1)
-80 kN · m - 5.75 kN x 1 + M = 0
M = (5.75x1 + 80) kN · m
X2 -
S.75 kN
S
X2 -
~
(c)
(2)
10 m, Fig.11- 7c:
+ jIFy = O; 5.75kN-15kN- 5kN/m(x2
v=
~+IM =
O;
-
SkN/m
c
5m) - V = 0
(15.75 - 5Xz) kN
A
(3)
B
Sm - - - -Sm
5.75 kN
34.2SkN
V(kN)
-80kN·m - 5.75kNXz + 15kN(Xz - 5m)
+5kN/m(Xz-5m) (
S
- 2- - 2-
15 kN
5 m < Xz
S)
>--- >---1
1 - - - - - X2
O;
-
(j..,_____~'i)M
V = 5.75 kN
~+IM =
t t t t tSkN/m
u
, ~
•
1 - -5 m - - l -5 m - I
(a)
< 5 m, Fig.11- 7b:
+ jIF.y = O·,
50 5
SHEAR AND MOMENT DIAGRAMS
5
X2 -
2
I x(m)
S.7S
m) + M = O
M = (-2.5xi2 + 15.75Xz + 92.5) kN · m
- 9.2S
(4)
Shear and Moment Diagrams. Equations 1 through 4 are plotted
in Fig.11- 7d.
M(kN·m)
-34.25
108.75
1----------~x(m)
(d)
Fig. 11- 7
506
CHAPTER
11
BE ND I NG
11. 2
Failure of this table occurred at the brace
support on its right side. If drawn, the
bending-moment diagram for the table
loading would indicate this to be the point
of maximum internal moment.
GRAPHICAL METHOD FOR
CONSTRUCTING SHEAR AND
MOMENT DIAGRAMS
In cases where a beam is subjected to several different loadings,
determining V and M as functions of x and then plotting these equations
can become quite tedious. In this section a simpler method for constructing
the shear and moment diagrams is discussed- a method based on two
differential relations, one that exists between the distributed load and
shear, and the other between the shear and moment.
Regions of Distributed Load. For purposes of generality,
consider the beam shown in Fig. 11- &t, which is subjected to an arbitrary
loading. A free-body diagram for a very small segment 6.x of the beam is
shown in Fig. 11-8b. Since this segment has been chosen at a position x
where there is no concentrated force or couple moment, the results to be
obtained will not apply at these points.
Notice that all the loadings shown on the segment act in their positive
directions according to the established sign convention, Fig. 11- 3. Also,
both the internal resultant shear and moment, acting on the right face of
the segment, must be changed by a small amount in order to keep the
segment in equilibrium. The distributed load, which is approximately
constant over 6.x, has been replaced by a resultant force w6.x that acts at
!(6.x) from the right side. Applying the equations of equilibrium to the
segment, we have
..w6x
IV
v = w(x)
F
r-1
I
I
I
I
I
--,
I
I
I
I
1I
I
2 (6x)
v
M
er
0
lJMHM
'------' v + 6 v
6x
Free-body diagram
of segment 6x
(b)
(a)
Fig.11- 8
11 .2
+ j!F.>'
= O·,
50 7
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND M OMENT DIAGRAMS
V + w 6.x - (V + 6. V)
= 0
6.V=w6.x
C+ IM0 = O;
-V 6.x - M - w 6.x[~(6.x)] + (M + 6.M)
6.M
= 0
= V 6.x + w!{6.x)2
Dividing by 6.x and taking the limit as 6.x---+ 0, the above two equations
become
uv ·:J
dx
-
slope of
shear diagram
at each poinl
dM
dx
slope of
moment diagram
at each point
(11-1)
distributed
load intensity
at each point
~
(11- 2)
shear
at each
point
Equation 11-1 states that at any point the slope of the shear diagram
equals the intensity of the distributed loading. For example, consider the
beam in Fig. l l-9a. The distri buted loading is negative and increases
from zero to w 8 . Knowing this provides a quick means for drawing the
shape of the shear diagram. It must be a curve that has a negative slope,
increasing from zero to - w8 . Specific slopes wA = 0, - we, - w 0 , and
- w 8 are shown in Fig. I l-9b.
In a similar manner, Eq. 11-2 states that at any point the slope of the
moment diagram is equal to the shear. Since the shear diagram in
Fig. ll-9b starts at +VA, decreases to zero, and then becomes negative
and decreases to -V8 , the moment diagram (or curve) will then have an
initial slope of + VA which decreases to zero, then the slope becomes
negative and decreases to - Va. Specific slopes VA, Ve, V0 , 0, and -Va
are shown in Fig.11-9c.
(a) A •
v
v...
_ o_B
c
w = negative increasing
slope = negative increasing
o
-we
/
-wo
(b)
V =positive decreasing
M
slope =;~"Tiecreasing-w
8
Ve
Vo
(c)
Fig. 11-9
0
508
CHAPTER
11
BENDING
wdx
Equations 11- 1and11- 2 may also be rewritten in the form dV =
and dM = Vdx. Since w dx and V dx represent differential areas under
the distributed loading and the shear diagram, we can then integrate
these areas between any two points C and Don the beam, Fig. 11- 9d,
and write
c
(d)
v
jwdx I
I av=
(11- 3)
area under
distributed loading
change in
shear
M
(11-4)
change in
moment
area under
shear diagram
(f)
Fig. 11-9 (cont.)
F
..
M
v
M+ 6.M
Equation 11- 3 states that the change in shear between C and D is equal
to the area under the distributed-loading curve between these two points,
Fig. 11- 9d. In this case the change is negative since the distributed load
acts downward. Similarly, from Eq. 11-4, the change in moment between
C and D, Fig. 11- 9[, is equal to the area under the shear diagram within
the region from C to D. Here the change is positive.
Regions of Concentrated Force and Moment. A free-body
diagram of a small segment of the beam in Fig. 11- 8a taken from under
the force is shown in Fig. 11- lOa. Here force equilibrium requires
+ t 2£,, = O;
V + F - (V +
a V)
=
0
(11- 5)
l-1h -1 v + av
(a)
M
v
~-~
Thus, when Facts upward on the beam, then the change in shear, d V , is
positive so the values of the shear on the shear diagram will "jump"
upward. Likewise, if F acts downward, the jump (d V) will be downward.
When the beam segment includes the couple moment M 0 , Fig.11-lOb,
then moment equilibrium requires the change in moment to be
C+ 2M0
=
O;
M + dM - M0
-
V dx - M
=
0
Letting dx "" 0, we get
I- ax -I v+ av
(b)
Fig.11-10
dM = M0
(11-6)
In this case, if Mo is applied clockwise, the change in moment, dM, is
positive so the moment diagram will "jump" upward. Likewise, when Mo
acts counterclockwise, the jump (dM) will be downward.
11.2
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT D IAGRAMS
PROCEDURE FOR ANALYSIS
The following procedure provides a method for constructing the shear
and moment diagrams for a beam based on the relations among
distributed load, shear, and moment.
Support Reactions.
• Determine the support reactions and resolve the forces act ing
on the beam into components that are perpendicular and
parallel to the beam's axis.
Shear Diagram.
• Establish the V and x axes and plot the known values of t he
shear at the two ends of the beam.
• Notice how the values of the distributed load vary along the beam,
such as positive increasing, negative increasing, etc., and realize that
each of these successive values indicates the way the shear diagram
will slope (dV/dx = w). Here w is positive when it acts upward.
Begin by sketching the slope at the end points.
• If a numerical value of the shear is to be determined at a point, one
can find this value either by using the method of sections and the
equation of force equilibrium, or by using /::,,. V =
w dx, which
states that the change in the shear between any two points is equal to
the area under the load diagram between the two points.
J
Moment Diagram.
• Establish the M and x axes and plot the known values of t he
moment at the ends of t he beam.
• Notice how the values of the shear diagram vary along the beam,
such as positive increasing, negative increasing, etc., and realize
that each of these successive values indicates the way the moment
diagram will slope (dM/dx = V). Begin by sketching the slope at
the end points.
• A t the point where the shear is zero, dM / dx = 0, and therefore
this will be a point of maximum or minimum moment.
• If a numerical value of the moment is to be determined at the
point, one can find this value either by using the method of
sections and the equation of moment equilibrium, or by using
l::..M =
dx, which states that the change in moment between
any two points is equal to the area under the shear diagram
between the two points.
• Since w must be integrated to obtain I::.. V, and V is integrated to
obtain M , then if w is a curve of degree n, V will be a curve of degree
n + 1 and M will be a curve of degree n + 2. For example, if w is
uniform, V will be linear and M will be parabolic.
Jv
5 09
510
I
CHAPTER
EXAMPLE
11
11.s
BE ND I NG
I
Draw the shear and moment diagrams for the beam shown in Fig.11- lla.
p
(a)
p
SOLUTION
Support Reactions. The reaction at the fixed support is shown on the
free-body diagram, Fig.11- llb.
Shear Diagram. The shear at each end of the beam is plotted first,
Fig. 11- llc. Since there is no distributed loading on the beam, the slope
of the shear diagram is zero as indicated. Note how the force P at the
center of the beam causes the shear diagram to jump downward an
amount P, since this force acts downward.
Moment Diagram. The moments at the ends of the beam are plotted,
Fig. 11- lld. Here the moment diagram consists of two sloping lines, one
with a slope of +2P and the other with a slope of +P.
The value of the moment in the center of the beam can be determined
by the method of sections, or from the area under the shear diagram. If
we choose the left half of the shear diagram,
Mlx=L
=
Mlx=O + l!..M
Mlx=L
=
-3PL + (2P)(L)
=
p
slope= 0
2P t --
-
p
(b)
w=O
v
-PL
-.--
downward force P
downward jump P
-i
~----P
.___ _....._,,,_ _ _ _ _ _
_,__~
x
(c)
M
V = positive constant
slope = positive constant
r---7-=:==:::=::::=-c;E~11:cis with
slope P
- 3PL
Begins with
slope 2P
(d}
Fig.11-11
11 .2
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND M OMENT DIAGRAMS
11~
EXAMPLE
Draw the shear and moment diagrams for the beam shown in Fig.11-12a.
Mo
1--L-
-l
L-
-1
(a)
SOLUTION
Support Reactions. The reactions are shown on the free-body diagram
in Fig. 11-12b.
The shear at each end is plotted first, Fig. 11-12c.
Since there is no distributed load on the beam, the shear diagram has
zero slope and is therefore a horizontal line.
Moment Diagram. The mome nt is zero at each end, Fig. 11-12d. The
moment diagram has a constant negative slope of - M0 /2L since this is
the shear in the beam at each point. However, here the couple moment
Mo causes a jump in the mo me nt diagram at the beam's center.
Shear Diagram.
'
t---
l .
·1-L- -•'
L \
Mo
(b)
2L
v
Mo
2L
•
"' =0
slope = O
i - - - - - r - - ---..- - ...-- x
(c)
clockwise moment M 0
positive jump M 0
M
V = negative constant
slope = negative constant
o/2
- Mo/2
(d)
Fig. 11-12
511
512
I
11
CHAPTER
EXAMPLE
11.1
BENDING
I
Draw the shear and moment diagrams for each of the beams shown in
Figs. 11- 13a and 11- 14a.
SOLUTION
3kN/m
11- - - -
Support Reactions. The reactions at the fixed support are shown on
each free-body diagram, Figs. ll- 13b and 11- 14b.
4m - - - (a)
l¥
3kN/m
"(ij 11
24kN·m
11111
j
(b)
V (kN)
w negative constant
V slope negative constant
12
Shear Diagram. The shear at each end point is plotted first,
Figs.11- 13c and 11- 14c. The distributed loading on each beam indicates
the slope of the shear diagram and thus produces the shapes shown.
Moment Diagram. The moment at each end point is plotted first,
Figs.11- 13d and 11- 14d. Various values of the shear at each point on the
beam indicate the slope of the moment diagram at the point. Notice how
this variation produces the curves shown.
NOTE: Observe how the degree of the curves from w to V to M increases
L.J.Jll-"11.---------=:::::.- x
(c)
V positive decreasing
M (kN ·m) M slope positive decreasing
by one due to the integration of dV = w dx and dM = Vdx. For
example, in Fig. 11- 14, the linear distributed load produces a parabolic
shear diagram and cubic moment diagram.
i-----r----:::::::::=~-:= x
Ends with
zero slope
- 24
(d)
Fig.11-13
u-- - - - 3 m - - - (a)
2kN/m
3kN~
(t
~
3kN·m
V (kN)
3
M(kN·m)
I
w negative decreasing
V slope negative decreasing
V positive decreasing
M slope positive decreasing
i----t;---:----::::::::=~- x
-3
(d)
Fig.11-14
(m)
11.2
-
EXAMPLE
51 3
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND MOMENT DIAGRAMS
11.8
-
Draw the shear and moment diagrams for the cantilever beam in Rig. 11- 15a.
(b)
(a)
IV = 0
IV negative constant
V slope = 0 V slope negative constant
SOLUTION
V(kN)
Support Reactions. The support reactions at the fixed support B
are shown in Fig.11- 15b.
Shear Diagram. The shear at the ends is plotted first, Fig. 11- lSc.
Notice how the shear diagram is constructed by following the slopes
defined by the loading w.
Moment Diagram. The moments at the ends of the beam are
plotted first , Fig. 11- 15d. Notice how the moment diagram is
constructed based on knowing its slope, which is equal to the shear at
each point. The moment at x = 2 m can be found from the area under
the shear diagram. We have
M l.r=2 m =
Mlx=O + 6.M
=
0 + [-2kN(2m)J
=
-4kN · m
-5
V negative constant
M slope negative constant
V negative increasing
M slope negative increasing
-4
- 11
(d)
Of course, this same value can be determined from the met!hod of
sections, Fig. 11- lSe.
V=2kN
t ) M=4kN·m
l -2011--il
(e)
Fig.11-15
514
CHAPTER
EXAMPLE
11
BENDING
11.9
-
-
Draw the shear and moment diagrams for the overhang beam in Fig.11- 16a.
4kN/m
~
I
4m ~-2m(a)
SOLUTION
Support Reactions. The support reactions are shown in Fig.11- 16b.
2m -1
1 - - -4 m
2kN
wJO(b\OkN
Vslope = 0
)
.
w negative constant
V (kN)
V slope negative constant
Shear Diagram. The shear at the ends is plotted first, Fig. 11- 16c. The
slopes are determined from the loading and from this the shear diagram
is constructed. Notice the positive jump of 10 kN at x = 4 m due to the
force reaction.
Moment Diagram. The moments at the ends are plotted first,
Fig. 11- 16d. Then following the behavior of the slope found from the
shear diagram, the moment diagram is constructed. The moment at x = 4 m
is found from the area under the shear diagram.
V negative (c)
V positive
constant
decreasing
M slope negative
M slope positive
constant
decreasing
Mlx=4m = Mlx=O + 6.M = O + (-2kN(4m)) = -8kN · m
We can also obtain this value by using the method of sections, as shown
in Fig. 11- 16e.
M (kN·m)
V= 2kN
-8
(d)
A__.....,....,....,....,...,. t ; M= 8 kN·m
1---- 4m - J
Fig.11-16
2kN
(e)
11.2
EXAMPLE
515
GRAPHICAL METHOD FOR CONSTRUCTING SHEAR AND M OMENT DIAGRAMS
11 .10
The shaft in Fig. 1l-17a is supported by a thrust bearing at A and a journal
bearing at B. Draw the shear and moment diagrams.
120 lb/ft
~
l201b/ft
t - -- -12 ft - - - - 1
(a)
SOLUTION
Support Reactions. The support reactions are shown in Fig.11- l7b.
12 ft
(b)
\
240 lb w nega tive
increasing
V slope negative
increasing
V(lb)
480 1b
Shear Diagram. As shown in Fig.11- 17c, the shears at the ends of the
beam are +240 lb and - 480 lb. The point where V = 0 must be located. 240'h n-'-'..J.
To do this we will use the method of sections. The free-body diagram of 0 '---'"ff!-- - - - '"l<::--nir--+12_ x {ft)
the left segme nt of the shaft, sectioned at an arbitrary position x, is shown
in Fig. 11-l7e. He re the intensity of the distributed load at x is w = lOx,
V positive
decreasing
- 480
which has been found by proportional triangles, i.e., 120/12 = w /x.
M slope positjve
Thus, for V = 0,
decreasing
V negative increasing
j
+ f ~F,. = O;
240 lb - ktOx)x
x
=
=
J
M{lb·ft) v~o
M slope ; ~0-r--...
0
6.93 ft
Moment Diagram. The moment diagram starts and ends at 0. The
maximum mo ment occurs at x = 6.93 ft , where the shear is equal to
zero,sincedM/dx = V = O, Fig.ll-17d. FromFig.11- 17e, we have
C+LM = O;
Mm••+ H(10)(6.93)] 6.93 {~(6.93) ) - 240(6.93)
M max
M slope negative increasing
o~-------.- x {ft)
6.93
(d)
=0
__
,_ x
= 1109lb·ft
Finally, notice how integration, first of the loading w , which is linear,
produces a shear diagram which is parabolic, and then a moment diagram
which is cubic.
NOTE: Now test yourself by covering over the shear and moment
diagrams in Examples 11.1 through 11.4, and see if you can construct
them based on the concepts discussed here.
12
-
---
:L lOx
v
i"
1\1
x
Ay= 240 lb
(e)
Fig. 11- 17
516
CHAPTER
11
BE ND I NG
PRELIMINARY PROBLEM
Pl l-1. In each case, the beam is subjected to the loadings
shown. Draw the free-body diagram of the beam, and sketch
the general shape of the shear and moment diagrams. The
loads and geometry are assumed to be known.
!
l l l l l
tl l l l l
l!:
~------lt
(a)
(e)
.h:~-----i:
(b)
(f)
(c)
(g)
(
.M.
(d)
(h)
Prob. Pll-1
11.2
517
GRAPHICAL METHOD FOR C ONSTRUCTING SHEAR AND M OMENT DIAGRAMS
FUNDAMENTAL PROBLEMS
In each case, express the shear and moment functions in
terms of x, and then draw the shear and moment diagrams
In each case, draw the shear and moment diagrams for the
beam.
for the beam.
Fll-5.
Fll-1.
Prob. Fll-S
Prob, 11-1
Fll-2.
Fll-6.
9kN
10 kN/m
10 kN/m
A~C~B
3m~
I
3m - - - - - - <
3m~
Pron. r 1 l-6
Fll-7.
Fll-3.
6001b
2 kip/ft
2001b;r1
""''(~
~x~
AL l l l l l I I I
I-
I
9ft
Fll-4.
12kN/m
,l•
6fl - 1 : . 3 f 1 + 3 f t- j
Fll-8.
20kN
3m-----l
Prob. r il-4
•
Pr -~·I !-7
Prob. l- .1-3
[-x-1
I
1-'rob. t' al-S
518
CHAPTER
11
BE ND I NG
PROBLEMS
11-1. Draw the shear and moment diagrams for the shaft
and determine the shear and moment throughout the shaft
as a function of x for 0 < x < 3 ft, 3 ft < x < 5 ft, and
5 ft < x < 6 ft. The bearings at A and B exert only vertical
reactions on the shaft.
*11-4. Express the shear and moment in terms of x for
0 < x < 3 m and 3 m < x < 4.5 m, and then draw the
shear and moment diagrams for the simply supported beam.
300N/m
500lb
~
800 lb
ft:
~ ~
1-x-13ft_j-2ft-~lft~-
__,;J,,_i!,h:-~~~~--i~~-::::::;;~B
1-3m -~1.5m~
0.Sft
Prob. 11-4
Prob. 11-1
11-2 Draw the shear and moment diagrams for the beam, and
determine the shear and moment in the beam as functions of x
forO < x < 4ft,4ft < x < lOft, andlOft < x < 14ft.
200 lb/ft
250 lb
250lb
!
I
11-5. Express the internal shear and moment in the
cantilevered beam as a function of x and then draw the
shear and moment diagrams.
150 lb/ft
A
x~A
1--- 4ft
B
·I - - -
6
ft~--4
Prob. 11-5
ft - -
Prob. 11-2
11-3. Draw the shear and moment diagrams for the beam,
and determine the shear and moment throughout the beam
as functions of x for 0 < x < 6 ft and 6 ft < x < 10 ft.
2 kip/ft
!
1500N
10 kip
I
!
•
1-x-
11-6. Draw the shear and moment diagrams for the shaft.
The bearings at A and B exert only vertical reactions on the
shaft. Also, express the shear and moment in the shaft as a
function of x within the region 125 mm< x < 725 mm.
~o kip·ft
i
800N
A
ll;:::::::::::I~~~==:Q
r--x
I- I- - - 600 mm - - - - - 1 - JI
4 ft
6 ft
Prob. 11-3
8
125mm
75mm
Prob. 11-6
11.2
519
GRAPHICAL METHOD FOR C ONSTRUCTING SHEAR AND M OMENT DIAGRAMS
11-7. Express the internal shear and moment in terms of x
for 0 < x < L/2 , and L/2 < x < L, and then draw the
shear and moment diagrams.
11-10. Draw the shear and moment diagrams for the
shaft. The bearings at A and D exert only vertical reactions
on the shaft.
A
n
E
c
B
D
r
'
Prob. 11-7
I
I
SO lb
i
351 b
110 lb
*11-8. Draw the shear and moment diagranis for the beam,
and determine the shear and moment throughout the beam
as functions of x for 0 :S x :S 6 ft and 6 ft < x < 9 ft.
4 k"p
·1
2 kip/fl
~·
.
~,_J
~20 kip·ft
Prob. 11- 10
11-11. The crane is used to support the engine, which has
a weight of 1200 lb. Draw the shear and moment diagrams
of the boom ABC when it is in the horizontal position.
,,,_Jl_,.J
B
Prob.11-8
11-9. rr the force applied to the handle of the load binder
is 50 lb, determine the tensions T1 and T 2 in each end of the
chain and then draw the shear and moment diagrams for
the arm ABC.
Prob. 11- 11
T1
*11-12.
beam.
c
B
!
I
SO lb
12in.---1
3in. -
Tz
Prob. 11-9
Draw the shear and moment diagrams for the
520
CHAPTER
11
BE ND I NG
11-13. Draw the shear and moment diagrams for the beam.
Mo
Mo
Mo
{M
(
f:,is
*11-16. A reinforced concrete pier is used to support the
stringers for a bridge deck. Draw the shear and moment
diagrams for the pier. Assume the columns at A and B exert
only vertical reactions on the pier.
60 kN
Prob.11-13
60kN
A
B
11-14. Draw the shear and moment diagrams for the beam.
Prob.11-16
2 kip/ft
! ! ! 1!~
~'=======~
~===~~;
; ~!~a~B
=::::;;=
30kip·ft
11-17. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam as
functions of x , where 4 ft < x < 10 ft.
A . .
1 - -5 ft - l - -5 ft - - - - 5
ft~
Prob.11-14
150 lb/ft
200 lb·ft
.______________~~!)
I
11-15. Members ABC and BD of the counter chair are
rigidly connected at B and the smooth collar at D is allowed
to move freely along the vertical post. Draw the shear and
moment diagrams for member ABC.
4 ft
x--1----_ _ 6 ft - - - 1
4ft - -
Prob.11-17
11-18. The industrial robot is held in the stationary position
shown. Draw the shear and moment diagrams of the arm ABC
if it is pin connected at A and connected to a hydraulic cylinder
(two-force member) BD. Assume the arm and grip have a
uniform weight of 15 lb/in. and support the load of 40 lb at C.
- - - - -50 in.- - - - - -1
1
1.5 ft
J
Prob.11-15
Prob.11-18
11.2
521
GRAPHICAL METHOD FOR C ONSTRUCTING SHEAR AND M OMENT DIAGRAMS
11- 19. Determine the placement distance a of the roller
support so that the largest absolute value of the moment is
a minimum. Draw the shear and moment diagrams for this
condition.
p
11-23. The 150-lb man sits in the center of the boat, which
has a uniform width and a weight per linear foot of 3 lb/ft.
Determine the maximum internal bending moment.Assume
that the water exerts a uniform distributed load upward on
the bottom of the boat.
p
1-~-i-~-1
t----- a - - ---1
Prob. 11-23
Prob. 11-19
*11-20. Draw the shear and moment diagrams for the beam.
*11-24. Draw the shear and moment diagrams for the
beam.
800 lb/fl
•
A
t - - - -- - ---.--...--.--.-..---.--.---1
B
800 lb/ft
,____ 8 ft
Prob. 11-20
'Ok"(jl I I ~ I
!
1-s f1-l-s
----1
Prob.11-24
11-21. Draw the shear and moment diagrams for the beam.
2 kip/fl
----1- --- 8 ft
2 kip/ ft
!} I I
f1-~s
11
11-25. The footing supports the load transmitted by
the two columns. Draw the shear and moment diagrams for
the footing if the soil pressure on the footing is assumed to
be uniform.
ft-I
Prob.11-21
11-22. Draw the shear and moment diagrams for the
overhanging beam.
Prob. 11-25
11-26.
Draw the shear and moment diagrams for the beam.
3 kip/ft
3 kip/fl
, +i
Ah\
-·
I 1 l]
I
1----12 ri-1-6 ft -I
Prob. 11-22
j
,;
B
1
L
,,ro----i-- 6ft__j
Prob.11-26
522
CHAPTER
11
BE ND I NG
11-27. Draw the shear and moment diagrams for the beam.
11-30. Draw the shear and moment diagrams for the beam.
2kip
200 lb/ft
~
A
1 - 6ft
,_ _ L
3
''
I---
A
--+----2L _ _ _ _ _,
9 ft
--1-
B
9 ft - - - 1
Prob.11-30
3
Prob.11-27
11-3L The support at A allows the beam to slide freely
along the vertical guide so that it cannot support a vertical
force. Draw the shear and moment diagrams for the beam.
*11-28. Draw the shear and moment diagrams for the beam.
JV
:211il lllll ~reys
I.1
A (
I
I
1--;-1-;-1-;-J
Prob.11-31
Prob.11-28
11-29. Draw the shear and moment diagrams for the beam.
*11-32. The smooth pin is supported by two leaves A and B
and subjected to a compressive load of 0.4 kN/m caused by
bar C. Determine the intensity of the distributed load w0 of
the leaves on the pin and draw the shear and moment
diagram for the pin.
18kN/m
12 kN/m
c
B
Prob.11-29
Prob.11-32
11.2
523
GRAPHICAL M ETHOD FOR C ONSTRUCTING SHEAR AND M OMENT DIAGRAMS
11- 33. The shaft is supported by a smooth thrust bearing
at A and smooth journal bearing at B. Draw the shear and
moment diagrams for the shaft.
11-37.
Draw the shear and moment diagrams for the beam.
50kN/m
50kN/m
400N·m
---£"l-----~
Al-4.5
-'--I m-+-1 mj
m---+----4.5 m --
---i
Prob. 11-37
900N
Prob. 11-33
11-34. Draw the shear and moment diagrams for the
cantilever beam.
11-38. The beam is used to support a uniform load along
CD due to the 6-kN weight of the crate. Also, the reaction at
the bearing support B can be assumed uniformly distributed
along its width. Draw the shear and moment diagrams for
the beam.
-
0.5 m'\ 0.75 m
2.75m - - -1-L......l.-2m - -I
Prob. 11-34
11-35.
Prob. 11-38
Draw the shear and moment diagrams for the beam.
400N/m
200 N/ m
11-39. Draw the shear and moment diagrams for the
double overhanging beam.
4001b
!
A
4001b
200 lb/h
!
Prob. 11-35
*11-36. Draw the shear and moment diagrams for the
rod. Only vertical reactions occur at its ends A and B.
~12 lb/in.
Prob. 11-39
*11-40. Draw the shear and moment diagrams for the
simply supported beam.
10 kN
15kN·m
I
I
9>n
A@
'Y-----36 in.------1
1441b
72 lb
Prob. 11-36
Prob. 11-40
524
CHAPTER
11
BE ND I NG
11-41. The compound beam is fixed at A , pin connected
at B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
*11-44. Draw the shear and moment diagrams for the
beam.
8 kip/ft
600N
400N/m
A
Jl l l l l i
B x
1 - - - - - - - -8 ft - - - - - - - -1
Prob.11-44
Prob.11-41
11-42. Draw the shear and moment diagrams for the
compound beam.
11-45. A short link at Bis used to connect beams AB and
BC to form the compound beam. Draw the shear and
moment diagrams for the beam if the supports at A and C
are considered fixed and pinned, respectively.
SkN/m
l l ! I I ! 1! l:f I l l
A
I
C
B
LD
2m ~-lm-l-lm ~
Prob.11-42
Prob.11-45
11-43. The compound beam is fixed at A , pin connected
at B, and supported by a roller at C. Draw the shear and
moment diagrams for the beam.
11-46. The truck is to be used to transport the concrete
column. If the column has a uniform weight of w (force/length),
determine the equal placement a of the supports from the ends
so that the absolute maximum bending moment in the column
is as small as possible. Also, draw the shear and moment
diagrams for the column.
1 - - - - -L - - - - - I
2kN
I-a-I
3kN/m
-r-3m~
Prob.11-43
C
Prob.11-46
11.3
11.3
BENDING DEFORMATION OF A STRAIGHT M EMBER
525
BENDING DEFORMATION OF A
STRAIGHT MEMBER
In this section, we will discuss the deformations that occur when a straight
prismatic beam, made of homogeneous material, is subjected to bending.
The discussion will be limited to beams having a cross-sectional area that
is symmetrical with respect to an axis, and the bending moment is applied
about an axis perpendicular to this axis of symmetry, as shown in Fig.11-18.
The behavior of members that have unsymmetrical cross sections, or are
made of several different materials, is based on similar observations and
will be discussed separately in later sections of this chapter.
Consider the undeformed bar in Fig. 11-19a, which has a square cross
section and is marked with horizontal and vertical grid lines. When a
bending moment is applied, it tends to distort these lines into the pattern
shown in Fig. 11-19b. Here the horizontal lines become curved, while the
vertical lines remain straight but undergo a rotation. The bending moment
causes the material within the bo11om portion of the bar to stretch and
the material within the top portion to compress. Consequently, between
these two regions there must be a surface, called the neutral surface, in
which horizontal fibers of the material will not undergo a change in
length, Fig. 11-18. As noted, we will refer to the z axis that lies along the
neutral surface as the neutral axis.
Axis or
symmetry
y
.I!-lZ>~;::';'
Neutral
Neutral surface
x
axis
Longitudinal
axis
Fig. 11- 18
Horizontal Lines
become curved
Vertical Imes re main
straight. yet rotate
Before deformation
After deformation
(b)
(a)
Fig.11-19
526
CHAPTER
11
BE ND I NG
Note the distortion of the lines due to
bending of this rubber bar. The top line
stretches, the bottom line compresses, and
the center line remains the same length.
Furthermore the vertical lines rotate and yet
remain straight.
From these observations we will make the following three assumptions
regarding the way the moment deforms the material. Ftrst, the
longitudinal axis, which lies within the neutral surface, Fig. 11- 20a, does
not experience any change in length. Rather the moment will tend to
deform the beam so that this line becomes a curve that lies in the vertical
plane of symmetry, Fig. 11-20b. Second, all cross sections of the beam
remain plane and perpendicular to the longitudinal axis during the
deformation. And third, the small lateral strains due to the Poisson effect
discussed in Sec. 3.6 will be neglected. In other words, the cross section in
Fig. 11- 19 retains its shape.
With the above assumptions, we will now consider how the bending
moment distorts a small element of the beam located a distance x along
the beam's length, Fig.11- 20. This element is shown in profile view in the
undeformed and deformed positions in Fig. 11- 21. Here the line segment
x
(a)
y
(:=·'
~
I
-;;:;:--;r1\
1G
zl
axis
neutral
surface
( b)
Fig.11-20
11.3
BENDING DEFORMATION OF A STRAIGHT M EMBER
0'
p
Before
de format ion
p
After
defom1ation
(a)
(b)
Fig. 11-21
Ax, located on the neutral surface, does not change its length, whereas
any line segment As, located at the arbitrary distance y above the neutral
surface, will contract and become As' after deformation. By definition,
the normal strain along As is determined from Eq. 7- 11, namely,
E
As' - As
= lim-- - ~O
As
Now let's represent this strain in terms of the location y of the segment
and the radius of curvature p of the longitudinal axis of the element.
Before deformation, 11s = 11x, Fig. 11-21a. After deformation, Ax bas a
radius of curvature p, with center of curvature at point O' , Fig. 11- 21b,
so that Ax = As = p/18. Also, since As' bas a radius of curvature of
p - y, then As' = (p - y)/18. Substituting these results into the above
equation, we get
E
. (p - y) AIJ - pAIJ
= hm - - -- - - A0-0
pAIJ
or
y
E
= -p
(11- 7)
5 27
528
CHAPTER
11
BENDING
- Emax
Normal strain distribution
Fig.11-22
Since 1/ p is constant at x, this important result, e = -y / p, indicates
that the longitudinal normal strain will vary linearly with y measured
from the neutral axis. A contraction (-e) will occur in fibers located
above the neutral axis ( +y), whereas elongation ( +e) will occur in fibers
located below the axis (-y). This variation in strain over the cross section
is shown in Fig.11-22. Here the maximum strain occurs at the outermost
fiber, located a distance of y = c from the neutral axis. Using Eq. 11- 7,
since Emax = c / p, then by division,
E
--=
Emax
-(yfp)
c/p
So that
E -(y)E
=
C
max
(11- 8)
This normal strain depends only on the assumptions made with regard
to the deformation.
11 .4
THE FLEXURE FORMULA
52 9
11.4 THE FLEXURE FORMULA
In this section, we will develop an equation that relates the stress
distribution within a straight beam to the bending moment acting on its
cross section. To do this we will assume that the material behav,e s in a
linear elastic manner, so that by Hooke's Jaw, a linear variation of normal
strain, Fig. 11- 23a, must result in a linear variation in normal stress,
Fig. 11- 23b. Hence, like the normal strain variation, a will vary from zero
at the member's neutral axis to a maximum value, amax• a distance c
farthest from the neutral axis. Because of the proportionality of triangles,
Fig.11- 23b, or by using Hooke's Jaw, a = Ee, and Eq.11- 8, we can write
y
Normal strain variation
(profile view)
(a)
y
(11- 9)
This equation describes the stress distribution over the cross-sectional
area. The sign convention established here is significant. For positive M,
which acts in the +z direction, positive values of y give negative values
for a, that is, a compressive stress, since it acts in the negative x direction.
Similarly, negative y values will give positive or tensile values for er.
Bending stress variation
(profile view)
(b)
Fig.11- 23
This wood specimen failed in bending due to its fibers being
crushed at its top and torn apart at its bottom.
530
CHAPTER
11
BE ND I NG
Location of Neutral Axis. To locate the position of the neutral
axis, we require the resultant fo rce produced by the stress distribution
acting over the cross-sectional area to be equal to zero. Noting that the
force dF = a dA acts on the arbitrary element dA in Fig.11- 24, we have
u
0 = ldF = iadA
U max
y
M
=
-a:ma x
C
x
1
ydA
A
Since amax/c is not equal to zero, then
Bending stress variation
(11- 10)
Fig.11-24
In other words, the first moment of the member's cross-sectional area
about the neutral axis must be zero. This condition can only be satisfied
if the neutral axis is also the horizontal centroidal axis for the cross
section.* Therefore, once the centroid for the member's cross-sectional
area is determined, the location of the neutral axis is known.
Bending Moment. We can determine the stress in the beam if we
require the moment M to be equal to the moment produced by the stress
distribution about the neutral axis. The moment of dF in Fig. 11- 24 is
dM = y dF. Since dF = a dA, using Eq. 11- 9, we have for the entire
cross section,
or
M
r
C }A
= Umax
y2dA
(11- 11)
*Recall that the location y for the centroid of an area is defined from the equation
y = j y dA/ j dA. If y dA = 0, then y = 0, and so the centroid lies on the reference
(neutral) axis. See Appendix A.
J
11 .4
The integral represents the moment of inertia of the cross-sectional
area about the neutral axis.* We will symbolize its value as I. Hence,
Eq.11-11 can be solved for CTmax and written as
(11- 12)
Here
CTmax =
M =
c =
I
=
the maximum normal stress in the member, which occl!lrs at a
point on the cross-sectional area farthest away from the
neutral axis
the resultant internal moment, determined from the method
of sections and the equations of equilibrium, and calculated
about the neutral axis of the cross section
perpendicular distance from the neutral axis to a point farthest
away from the neutral axis. This is where CTmax acts.
moment of inertia of the cross-sectional area about the
neutral axis
Since CTmax/c = -u/y, Eq. 11-9, the normal stress at any distance y
can be determined from an equation similar to Eq.11-12. We have
(11- 13)
Either of the above two equations is often referred to as the flexure
formula. Although we have assumed that the member is prismatic, we
can conservatively also use the flexure formula to determine the normal
stress in members that have a slight taper. For example, using a
mathematical analysis based on the theory of elasticity, a member having
a rectangular cross section and a length that is tapered 15° will have an
actual maximum normal stress that is about 5.4o/o less than that calculated
using the flexure formula.
*See Appendix A for a discussion on how to determine the moment of inertia for various
shapes.
THE FLEXURE FORMULA
5 31
532
CHAPTER
11
BENDING
IMPORTANT POINTS
• The cross section of a straight beam remains plane when the beam deforms due to bending. This causes
tensile stress on one portion of the cross section and compressive stress on the other portion. In between
these portions, there exists the neutral axis which is subjected to zero stress.
• Due to the deformation, the longitudinal strain varies linearly from zero at the neutral axis to a maximum at
the outer fibers of the beam. Provided the material is homogeneous and linear elastic, then the stress also
varies in a linear fashion over the cross section.
• Since there is no resultant normal force on the cross section, then the neutral axis must pass through the
centroid of the cross-sectional area.
• The flexure formula is based on the requirement that the internal moment on the cross section is equal to the
moment produced by the normal stress distribution about the neutral axis.
PROCEDURE FOR ANALYSIS
In order to apply the flexure formula, the following procedure is suggested.
Internal Moment.
• Section the member at the point where the bending or normal stress is to be determined, and obtain the
internal moment M at the section. The centroidal or neutral axis for the cross section must be known,
since M must be calculated about this axis.
• If the absolute maximum bending stress is to be determined, then draw the moment diagram in order to
determine the maximum moment in the member.
Section Property.
• Determine the moment of inertia of the cross-sectional area about the neutral axis. Methods used for its
calculation are discussed in Appendix A, and a table listing values of I for several common shapes is given on
the inside front cover.
Normal Stress.
• Specify the location y, measured perpendicular to the neutral axis to the point where the normal stress is
to be determined. Then apply the equation <T = -My/I, or if the maximum bending stress is to be
calculated, use <Tmax = Mc/ I. When substituting the data, make sure the units are consistent.
• The stress acts in a direction such that the force it creates at the point contributes a moment about the
neutral axis that is in the same direction as the internal moment M . In this manner the stress distribution
acting over the entire cross section can be sketched, or a volume element of the material can be isolated and
used to graphically represent the normal stress acting at the point, see Fig.11-24.
11.4
I
EXAMPLE
11 .11
533
THE FLEXURE FORMULA
I
The simply supported beam in Fig. 11- 25a has the cross-sectional area
shown in Fig. 11- 25b. Determine the absolute maximum bending
stress in the beam and draw the stress distribution over the cross
section at this location. Also, what is the stress at point B?
M (kN·m}
SkN/ m
1
!~r~1~r1~1~r1~1~11~1~r
22.5
a•
""-- - - - - - - - - - - - - -""'--- x (m)
3
(a)
6
(c)
SOLUTION
20mm
Maximum Internal Moment. The maximum internal moment in
the beam, M = 22.5 kN · m, occurs at the center, as indicated on the
moment diagram, Fig. 11- 25c.
By reasons of symmetry, the neutral axis passes
through the centroid Cat the rnidheight of the beam, Fig. 11- 25b. The
area is subdivided into the three parts shown, and the moment of
inertia of each part is calculated about the neutral axis using the
parallel-axis theorem. (See Eq. A- 5 of Appendix A.) Choosing to
work in meters, we have
Section Property.
l _____,
-I s71
1
v-c
150mm
N -~~-i•1--~--'J'--A
1
20mm 150mm
I
J
20 mm == c:::==::'.:.:::==::::r-'-
1
-
250mm -
(b)
I = l(l + Ad2)
=
1~ (0.25 m)(0.020 m) + (0.25 m)(0.020 m)(0.160 m)
+ [ ~ (0.020 m)(0.300 m)
1
3
2[
3
=
ll.2MP~
]
~
]
301.3(10- 6) m4
Mc
<Tmax
2
= -
I
;
<Tmax
=
22.5(103) N · m(0.170 m)
-6
301.3(10 ) m4
=
12.7 MPa
Ans.
A three-dimensional view of the stress distribution is shown in
Fig. 11- 25d. Specifically, at point B , y 8 = 150 mm, and so as shown
in Fig. 11- 25d,
My8
<Ts = - ,
1
22.5(103) N · m(0.150 m)
<Ts = 301.3(10- 6) m4
12.7 MPa
(d}
=
-11.2 MPa Ans.
Fig.11-25
11.2MPa
534
CHAPTER
-EXAMPLE
11
BENDING
11.12
The beam shown in Fig.11- 26a has a cross-sectional area in the shape
of a channel, Fig. ll- 26b. Determine the maximum bending stress that
occurs in the beam at section a- a.
2.6kN
~12
a
SOLUTION
- - -2m - - -1-
a
lm
(a)
L
1- 2somm l
y=59.09mm
~
N t
15mm-
C
]~
Wmo1 I
200mm
-
- 15mm j
(b)
_
Y=
2.4kN
1.0 kN
v
0.05909 m
Here the beam's support reactions do not have
to be determined. Instead, by the method of sections, the segment to
the left of section a-a can be used, Fig. 11- 26c. It is important that
the resultant internal axial force N passes through the centroid of the
cross section. Also, realize that the resultant internal moment must be
calculated about the beam's neutral axis at section a- a.
To find the location of the neutral axis, the cross-sectional area is
subdivided into three composite parts as shown in Fig. ll- 26b. Using
Eq. A- 2 of Appendix A , we have
2yA
2(0.100 m](0.200 m)(0.015 m) + (0.010 m](0.02 m)(0.250 m)
2A =
2(0.200 m)(0.015 m) + 0.020 m(0.250 m)
Internal Moment.
0.05909 m = 59.09 mm
This dimension is shown in Fig.11- 26c.
Applying the moment equation of equilibrium about the neutral
axis, we have
C+ 2MNA = 0; 2.4 kN(2 m) + 1.0 kN(0.05909 m) - M = 0
M = 4.859 kN · m
Section Property. The moment of inertia of the cross-sectional area
about the neutral axis is determined using I = L (l + Ad2) applied to each
of the three composite parts of the area. Working in meters, we have
=
M
ri"-----~=::=~=-.=.~ N
c
i - - - 2 m - - -1
(c)
Fig.11-26
I = [
~
(0.250 m)(0.020 m)3 + (0.250 m)(0.020 m)(0.05909 m 1
+ 2[
~
(0.015 m)(0.200 m) 3 + (0.015 m)(0.200 m)(0.100 m 1
=
0.010 m) 2 ]
0.05909 m) 2 ]
42.26(10-6) m4
The maximum bending stress occurs at
points farthest away from the neutral axis. This is at the bottom of the
beam, c = 0.200 m - 0.05909 m = 0.1409 m. Here the stress is
compressive. Thus,
Mc
4.859(103) N · m(0.1409 m)
Ans.
amax = I
=
= 16.2 MPa (C)
42.26(l0-6) m4
Show that at the top of the beam the bending stress is a' = 6.79 MPa.
Maximum Bending Stress.
The normal force of N = 1 kN and shear force V = 2.4 kN
will also contribute additional stress on the cross section. The
superposition of all these effects will be discussed in Chapter 13.
NOTE:
11 .4
-
EXAMPLE
THE FLEXURE FORMULA
11 .13
-
The member having a rectangular cross section, Fig. 11- 27a, is
designed to resist a moment of 40 N · m. In order to increase its
strength and rigidity, it is proposed that two small ribs be added at its
bottom, Fig. 11- 27b. Determine the maximum normal stress in the
member for both cases.
SOLUTION
Clearly the neutral axis is at the center of the cross
section, Fig.11- 27a, soy = c = 15 mm = 0.015 m. Thus,
Without Ribs.
1 3
bh = ~ (0.060 m)(0.030 m) 3 = 0.135(10- 6) m4
12
Therefore the maximum normal stress is
Mc
(40 N · m)(0.015 m)
CTmax = =
= 4.44 MPa
I
0.135(10- 6) m4
(a)
1
I =
Ans.
From Fig. 11- 27b, segmenting the area into the large
main rectangle and the bottom two rectangles (ribs), the location y of
the centroid and the neutral axis is determined as follows:
With Ribs.
I-A
y
y =--
Fig.11-27
IA
(0.015 m](0.030 m)(0.060 m) + 2(0.0325 m](0.005 m)(0.010 m)
(0.03 m)(0.060 m) + 2(0.005 m)(0.010 m)
= 0.01592 m
This value does not represent c. Instead
-
c = 0.035 m - 0.01592 m = 0.01908 m
Using the parallel-axis theorem, the moment of inertia about the
neutral axis is
I = [
~
(0.060 m)(0.030 m) 3 + (0.060 m)(0.030 m)(0.01592 m
1
+ 2[
(b)
- 0.015 m) 2 ]
~
(0.010 m)(0.005 m) 3 + (0.010 m)(0.005 m)(0.0325 m 1
0.01592 m) 2 ]
= 0.1642(10- 6) m4
Therefore, the maximum normal stress is
Mc
40 N · m(0.01908 m)
CTmax = =
= 4.65 MPa
I
0.1642(10- 6) m4
Ans.
NOTE: This surprising result indicates that the addition of the ribs to
the cross section will increase the maximum normal stress rather than
decrease it, and for this reason, the ribs should be omitted.
535
536
CHAPT ER
11
BEND I NG
PRELIMINARY PROBLEMS
Pll-2. Determine the moment of inertia of the cross
section about the neutral axis.
Pll-4. In each case, show how the bending stress acts on a
differential volume element located at point A and point B.
p
A
l
I
i
:a
0
B
(a)
0.2m
-tN
I
I A
0.1 m
I
0.1 m
l_
0.2m
A
M
(!
-
0.1 m
0.2m -
a
B
M
!)
~
I~
( )
(b)
Prob. Pll- 2
Prob.Pll-4
Pll-3. Determine the location of the centroid, y , and
the moment of inertia of the cross section about the
neutral axis.
Pll- 5. Sketch the bending stress distribution over each
cross section.
O.lm
I
I I
0.3m
L:---+---------+-A
t
y
0.1 m
l_ .__~~~_,c_~--'l -0.2
m-I
Prob. Pll- 3
(a)
(b)
Prob.Pll-5
11 .4
THE FLEXURE FORMULA
537
FUNDAMENTAL PROBLEMS
Fll-9. If the beam is subjected to a bending moment of
M = 20 kN · m, determine the maximum bending stress in
the beam.
Fll-12. If the beam is subjected to a bending moment of
M = 10 kN · m, determine the bending stress in the beam
at points A and B, and sketch the results on a differential
element at each of these points.
Prob.Fll-9
Fll-10. If the beam is subjected to a bending moment of
M = 50 kN · m, sketch the bending stress distribution over
the beam's cross section.
B
Prob. Fll-12
Fll-13. If the beam is subjected to a bending moment of
M = 5 kN · m, determine the bending stress developed at
point A and sketch the result on a differential element at
this point.
Prob.Fll-10
~~50mm
Fll-11. If the beam is subjected to a bending moment of
M = 50 kN · m, determine the maximum bending stress in
the beam.
~
150mm
0
I
300mm
150mm
M
20mm
D
20mm
~
Prob.Fll-11
Prob. Fll-13
538
CHAPTER
11
BE ND I NG
PROBLEMS
11-47. An A-36 steel strip has an allowable bending stress
of 165 MPa. If it is rolled up, determine the smallest radius r
of the spool if the strip has a width of 10 mm and a thickness
of 1.5 mm. Also, find the corresponding maximum internal
moment developed in the strip.
11- 50. The beam is constructed from four pieces of wood,
glued together as shown. If M = 10 kip · ft, determine the
maximum bending stress in the beam. Sketch a threedimensional view of the stress distribution acting over the
cross section.
ll- 5L The beam is constructed from four pieces of wood,
glued together as shown. If M = 10 kip · ft, determine the
resultant force this moment exerts on the top and bottom
boards of the beam.
Prob.11-47
*11-48. Determine the moment M that will produce a
maximum stress of 10 ksi on the cross section.
11-49. Determine the maximum tensile and compressive
bending stress in the beam if it is subjected to a moment of
M = 4 kip· ft.
Probs. 11- 50/51
*11-52. The beam is made from three boards nailed
together as shown. If the moment acting on the cross section
is M = 600 N · m, determine the maximum bending stress in
the beam. Sketch a three-dimensional view of the stress
distribution and cover the cross section.
11- 53. The beam is made from three boards nailed
together as shown. If the moment acting on the cross section
is M = 600 N · m, determine the resultant force the bending
stress produces on the top board.
!$.~ in.:41 i -'J-in,,...~·~===~
I
c
Q«
0.5 in '
25mm
3 in.
0
150mm
LI-J D
l- o.s in.
Probs.11-48/49
~1
/\I
200 mm M = 600 N·m
; 1V
20mm
Probs. 11- 52/53
11.4
11-54. If the built-up beam is subjected to an internal
moment of M = 75 kN · m. determine the maximum tensile
and compressive stress acting in the beam.
11-55. If the built-up beam is subj ected to a n internal
mome nt of M = 75 kN · m, determine the amount of this
inte rnal mome nt resisted by plate A.
T HE FLEXURE FORMULA
539
11-58. The beam is made from three boards nailed
together as shown. If the moment acting on the cross section
is M = 1 kip · ft, determine the maximum bending stress in
the beam. Sketch a three-dimensional view of the stress
distribution acting over the cross section.
11-59. If M = I kip· ft, determine the result ant force the
bending stresses produce on the top board A of th e beam.
300mm
1.5 in.
Probs. 11-58/59
Probs. 11-54155
•U -56. The beam is subjected to a moment M. Determine the
percentage of this moment that is resisted by the stresses acting
on both the top and bottom boards of the beam.
11-57. Dete rmine the moment M that sho uld be applied to
the beam in order to create a compressive stress at point D
of u 0 = 10 MPa. Also sketch the stress distribution acting
over the cross section and calculate th e maximum stress
developed in the beam.
-U-00- The beam is subjected to a moment of IS kip· ft.
D etermine the resultant force the bending stress produces
on the top Oange A and bottom flange B. Also calculate the
maximum bending stress developed in the beam.
11-61. The beam is subjected to a mome nt of 15 kip· ft.
Dete rmine the pe rce ntage of this mome nt that is resisted by
the web D of the beam.
M = 15 kip·ft
Probs. 11-56157
Probs.11-00/61
540
CHAPTER
11
BENDIN G
11-62. The beam is subjected to a moment of M = 40 kN · m.
Determine the bending stress at points A and 8. Sketch the
results on a volume element acting at each of these points.
11-65. A shaft is made of a polymer having an elliptical
cross section. If it resists an internal moment of
M = 50 N · m. determine the maximum bending stress in
the material (a) using the flexure formula, where
I,= k11(0.08 m)(0.04 m) 3, (b) using integration. Sketch a
three-dimensional view of the stress distribution acting over
the cross-sectional area. Here Ix = }11(0.08 m)(0.04 m) 3 .
11-66. Solve Prob. 11-65 if the moment M = 50 N · m is
applied about the y axis instead of the x axis. Herc
11 = '7T(0.04 m)(0.08 m)3 .
~M=40kN·m
!
50mm
y
y2
Prob. 11-62
z2
-- + -- = I
(4())2 (80)2
11-63. The steel shaft bas a diameter of2 in. It is supported
on smooth journal bearings A and 8, which exert only
vertical reactions on the shaft. Determine the absolute
maximum bending stress in the shaft if it is subjected to the
pulley loadings shown.
A
x
B
[]
fl
Probs. 11-65166
l20 in.+ 20 in. , 70 in.i-20 in.-l
500 Ib
300lb
500 lb
Prob. 11-63
*ll-64. ·n1c beam is made of steel that has an allowable
stress of Uauow = 24 ksi. Determine the larges t internal
moment the beam can resist if the moment is applied
(a) about the z axis, (b) about the y axis.
y
......_
0.25
in.
11-67. The shaft is supported by smooth journal bearings
at A and 8 that only exert vertical reactions on the shaft. If
d = 90 mm, determine the absolute maximum bending
stress in the beam, and sketch the stress distribution acting
over the cross section.
*ll-68. The shaft is supported by smooth journal bearings
at A and 8 that only exert vertical reactions on the shaft.
Determine its smallest diameter d if the allowable bending
stress is Uallow = 180 MPa.
~.
~~~~~~~~-~1~
2kN/m
J ll l llll l ~, ~
1 - - -- -
Prob. 11-64
3m -
-
-
-+-
1.5 m
Probs.11-67/68
J
11 .4
11-69. The axle of the freight car is subjected to a wheel
loading of 20 kip. If it is supported by two journal bearings at
C and D, determine the maximum bending stress developed
at the center of the axle, where the diameter is S.S in.
541
THE FLEXURE FORMULA
*11-72. Determine the absolute maximum bending stress
in the LS-in.-diameter shaft. The shaft is supported by a
thrust bearing al A and a journal bearing al B.
11-73. Determine the smallest allowable diameter of the
shaft. The shaft is supported by a thrust bearing at A and a
journal bearing at B. The allowable bending stress is
u a11ow = 22 ksi.
4001b
A
1-- - - 6 0 in.- ---·i---i
10 in.
20kip
3001b
8
10 in.
20 kip
Prob. 11-69
11-70. The strut on the utility pole supports the cable having
a weight of 600 lb. Determine the absolute maximum bending
stress in the strul if A, 8, and C arc asswned to be pinned.
----4 ft----- 2 rt-- 1
=o
2 in.
--l I,.·'-----,.......::•~----,--'
1_4 in.
1
8
1.5 ft
Probs. 11-72/73
11-74. The pin is used 10 connect th e three links together.
Due to wear, the load is distributed over the top and bottom
of the pin as shown on the Cree-body diagram. If the
diameter of the pin is 0.40 in., determine the maximum
bending stress on the cross-sectional area at the center
section a-a. For the solution it is first necessary to determine
the load intensities w 1 and w2•
8001b
t
6001b
Prob. 11-70
"'2
t-1 in.-t
11-7L The boat has a weight of 2300 lb and a center of
gravity at G. If it rests on the trailer at the smooth contact A
and can be considered pinned at B. determine the absolute
maximum bending stress developed in the main strut of
the trailer which is pinned al C. Consider the strut to be a
box-beam having the dimensions shown.
H in.-t
0
"'2
a
ro.40in.
•
1-1.5 in.-1
'
4001b
400 1b
Prob. 11-'74
11-75. The shaft is supported by a thrust bearing at A
and journal bearing at D. If the shaft has the cross section
shown, determine the absolute maximum bending stress in
the shaft.
1.75 in.
1 -1
3
in~~Il.75 in.
1.5 in.
Prob. 11-71
B
I0.75 m
;:==i v40 n~ mm
--1.S m1--r---1
3kN
3 kN
Prob. 11-75
542
CHAPTER
11
BE ND I NG
*11-76. If the intensity of the load w = 15 kN /m,
determine the absolute maximum tensile and compressive
stress in the beam.
11-81. If the compound beam in Prob. 11-42 has a square
cross section of side length a, determine the minimum val ue
of a if the allowable bending stress is uauow = 150 MPa.
11-77. If the allowable bending stress is u allow = 150 MPa,
determine the maximum intensity w of the uniform
distributed load.
11-82. If t he beam in Prob. 11-28 has a rectangular cross
section with a width band a height h, determine the absolute
maximum bending stress in the beam.
lf l l l ill l l ii l l i f1_
11-83. Determine the absolute maximum bending stress
in the SO-mm-diameter shaft which is subjected to the
concentrated forces. There is a journal bearing at A and a
thrust bearing at B.
IV
I
- - - -6m
*11-84. D etermine, to the nearest millimeter, the smallest
allowable diameter of the shaft which is subjected to the
concentrated forces. There is a journal bearing at A and a
thrust bearing at B. The allowable bending stress is
Uauow = 150 MPa.
81mm
l-1
150mm
Probs.11-76n7
11-78. The beam is subjected to the triangular distributed
load with a maximum intensity of w0 = 300 lb/ft. If the
allowable bending stress is u allow = 1.40 ksi, determine the
required dimension b of its cross section to the nearest
in. Assume the support at A is a pin and B is a roller.
l
11-79. The beam has a rectangular cross section with b =4
in. Determine the largest maximum intensity w0 of the
triangular distributed loads that can be supported if the
allowable bending stress is u allow= 1.40 ksi.
-
6ft
-
Probs. 11-78n9
*11-80. Determine the absolute maximum bending stress
in the beam. Each segment has a rectangular cross section
with a base of 4 in. and height of 12 in.
12 kip
ol
A
B
I
9 ft
i--
B
0.5 m- - - - 0.4 m
-i---
12 kN
20kN
Probs. 11-83184
'
Ic
14 kip
!
Ai-I- -4.5 ft - - - - -4.5 ft- -
~
- 3ft-
Prob.11-80
0.6 m - - - 1
11-85. Determine the absolute maximum bending stress
in the beam, assuming that the support at B exerts a
uniformly distributed reaction on the beam. The cross
section is rectangular with a base of 3 in. and height of 6 in.
6ft
2 kip/ft
A
- 6ft-I
Prob.11-85
B
3 ft -J
11.4
543
THE FLEXURE FORMULA
11-86. Determine the absolute maximum bending stress
in the 2-in.-diameter shaft. There is a journal bearing at A
and a thrust bearing at B.
11-90. If the beam in Prob. 11-19 has a rectangular cross
section with a width of 8 in. and a height of 16 in., determine
the absolute maximum bending stress in the beam.
11-87. Determine the smallest diameter of the shaft to
the nearest kin. There is a journal bearing at A and a thrust
bearing at B. The allowable bending stress is uallow = 22 ksi.
11-9L The simply supported truss is subjected to the
central distributed load. Neglect the effect of the diagonal
lacing and determine the absolute maximum bending stress
in the truss. The top member is a pipe having an outer
diameter of 1 in. and thickness of ;36 in., and the bottom
member is a solid rod having a diameter of! in.
900 Jb
_@
B
5.75 in. J
100 lb/ft
11
6ft
6f1
- o
-~6ft~
Prob.11-91
*11-92. If d = 450 mm, determine the absolute maximum
bending stress in the overhanging beam.
Probs. 11-86/87
11-93. If the allowable bending stress is u allow = 6 MPa,
determine the minimum dimension d of the beam's
cross-sectional area to the nearest mm.
*11-88. A log that is 2 ft in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress is u allow= 8 ksi, determine the
required width b and height h of the beam that will support
the largest load possible. What is this load?
11-89. A log that is 2 ft in diameter is to be cut into a
rectangular section for use as a simply supported beam. If
the allowable bending stress is u allow= 8 ksi, determine the
largest load P that can be supported if the width of the
beam is b = 8 in.
I
\
25mm )25mm
12kN
8 kN/m
l::lllllllll :
8
- - - 4 m - ---1- - 2
1
m
-1
'
l-=h- 1
- 2ft -
75mm~C
1
+
75mn~ I
_[
Probs. 11-92/93
11-94. The beam has a rectangular cross section as shown.
Determine the largest intensity w of the uniform distributed
load so that the bending stress in the beam does not exceed
u max= 10 MPa.
1r r r r r r r r r r r r r
0
l----8ft- -ll - - -8ft -----l
Probs. 11-88189
d
I
p
0
w
-11
11-95. The beam has the rectangular cross section shown.
If w = 1 kN/m, determine the maximum bending stress in
the beam. Sketch the stress distribution acting over the
cross section.
h
I •
!
125 mm
!11~
.._I-----1~1----~-----"I
I \
f t
- -2
m-J--2m --~2 m Probs. 11-94/95
-
0
( 150 mm
544
CHAPTER
11
BENDING
11.5
y
Axis of symmetry
UNSYMMETRIC BENDING
When developing the flexure formula, we required the cross-sectional
area to be symmetric about an axis perpendicular to the neutral axis
and the resultant moment M act along the neutral axis. Such is the
case for the "T " sections shown in Fig. 11- 28. In this section we will
show how to apply the flexure formula either to a beam having a
cross-sectional area of any shape or to a beam supporting a moment
that acts in any direction.
Moment Applied About Principal Axis. Consider the beam's
y
Axis of symmetry
cross section to have the unsymmetrical shape shown in Fig. 11- 29a. As
in Sec. 11.4, the right-handed x, y, z coordinate system is established such
that the origin is located at the centroid C on the cross section, and the
resultant internal moment M acts along the +z axis. It is required that
the stress distribution acting over the entire cross-sectional area have a
zero force resultant. Also, the moment of the stress distribution about
they axis must be zero, and the moment about the z axis must equal M .
These three conditions can be expressed mathematically by considering
the force acting on the differential element dA located at (0, y, z),
Fig. 11- 29a. Since this force is dF = u dA , we have
FR = IF_,;
0 = -iudA
(11- 14)
0 = - i
(11- 15)
Fig.11-28
(MR)y
=
lMy;
(MR)z
=
lMz;
M
=
zudA
(11- 16)
iyudA
y
dF=udA
x
Bending-stress distribution
(profile view)
(b)
(a)
Fig.11-29
11.5
UNSYMMETRIC BENDING
545
As shown in Sec. 11.4, Eq. 11- 14 is satisfied since the z axis passes
through the centroid of the area. Also, since the z axis represents the
neutral axis for the cross section, the normal stress will vary linearly from
zero at the neutral axis to a maximum at IYI = c, Fig.11- 29b. Hence the
stress distribution is defined by CT = -(y / c)CTmax . When this equation is
substituted into Eq. 11- 16 and integrated, it leads to the flexure formula
CTmax = Mc/I. When it is substituted into Eq.11- 15, we get
0 =
-CT,max
C
1
YZ dA
A
Z- sectioned members are often used in
light-gage metal building construction
to support roofs. To design them to
support bending loads, it is necessary to
determine their principal axes of inertia.
which requires
This integral is called the product of inertia for the area. As indicated
in Appendix A , it will indeed be zero provided the y and z axes are
chosen as principal axes of inertia for the area. For an arbitrarily shaped
area, such as the one in Fig. 11- 29a, the orientation of the principal axes
can always be determined, using the inertia transformation equations as
explained in Appendix A, Sec. A.4. If the area has an axis of symmetry,
however, the principal axes can easily be established since they will
always be oriented along the axis of symmetry and perpendicular to it.
For example, consider the members shown in Fig. 11- 30. In each of
these cases, y and z represent the principal axes of inertia for the cross
section. In Fig. 11- 30a the principal axes are located by symmetry, and
in Figs. 11- 30b and 11- 30c their orientation is determined using the
methods of Appendix A. Since M is applied only about one of the
principal axes (the z axis), the stress distribution has a linear variation,
and is determined from the flexure formula, CT = -My/ lz, as shown for
each case.
y
y
y
z
z
(a)
(b)
Fig.11-30
(c)
546
CHAPTER
11
BENDING
y
Moment Arbitrarily Applied. Sometimes a member may be
(a)
loaded such that M does not act about one of the principal axes of the
cross section. When this occurs, the moment should first be resolved into
components directed along the principal axes, then the flexure formula
can be used to determine the normal stress caused by each moment
component. Fmally, using the principle of superposition, the resultant
normal stress at the point can be determined.
To formalize this procedure, consider the beam to have a rectangular
cross section and to be subjected to the moment M,Fig.11- 31a, where M
makes an angle 6 with the maximum principal z axis, i.e., the axis of
maximum moment of inertia for the cross section. We will assume 6 is
positive when it is directed from the +z axis towards the +y axis.
Resolving M into components, we have Mz = M cos 6 and My = M sin 6,
Figs. 11- 31b and 11- 31c. The normal-stress distributions that produce M
and its components M z and M y are shown in Figs. 11- 31d, 11- 31e, and
11- 31/, where it is assumed that (o:r)max > (a"x)max· By inspection, the
maximum tensile and compressive stresses ((o:r)max + (a'x)max] occur at
two opposite corners of the cross section, Fig.11- 3ld.
Applying the flexure formula to each moment component in Figs. 11- 31b
and 11- 31c, and adding the results algebraically, the resultant normal
stress at any point on the cross section, Fig.11- 3ld, is therefore
II
y
(11- 17)
x
(b)
+
Here,
a = the normal stress at the point. Tensile stress is positive and
compressive stress is negative.
y
y, z
=
the coordinates of the point measured from a right-handed
coordinate system, x, y, z, having their origin at the centroid of
the cross-sectional area. The x axis is directed outward from the
cross section and t!he y and z axes represent, respectively, the
principal axes of minimum and maximum moment of inertia
for the area.
Mz, My
=
the resultant internal moment components directed along the
maximum z and minimum y principal axes. They are positive if
directed along the + z and +y axes, otherwise they are negative.
Or, stated another way, My = M sin 6 and Mz = M cos 6, where
6 is measured positive from the +z axis towards the +y axis.
Iv ly
=
the maximum and minimum principal moments of inertia
calculated about the z and y axes, respectively. See Appendix A.
(c)
Fig.11-31
11.5
Orientation of the Neutral Axis. The equation defining the
neutral axis, and its inclination a , Fig. 11- 31d, can be determined by
applying Eq. 11- 17 to a pointy, z where a = 0, since by definition no
normal stress acts on the neutral axis. We have
Since Mz
=
M cos 8 and My
=
y
M sin 8, then
= (
1
z tan
ly
o)z
UNSYMMETRIC BENDING
y
[(u" max -
((u,)m•x + (u~)m•x]
(11- 18)
(d)
II
Since the slope of this line is tan a
= y/
z, then
lz
tan a = -tan 8
ly
(11- 19)
(u x)max
(e)
IMPORTANT POINTS
+
• The flexure formula can be applied only when bending occurs
about axes that represent the principal axes of inertia for the
cross section. These axes have their origin at the centroid and
are oriented along an axis of symmetry, if there is one, and
perpendicular to it.
• If the moment is applied about some arbitrary axis, then the
moment must be resolved into components along each of the
principal axes, and the stress at a point is determined by
superposition of the stress caused by each of the moment
components.
(f)
Fig. 11-31 (cont.)
547
548
I
CHAPTER
EXAMPLE
11
BENDING
11.14 1
The rectangular cross section shown in Fig.11- 32a is subjected to a bending
moment of M = 12 kN · m. Determine the normal stress developed at each
corner of the section, and specify the orientation of the neutral axis.
SOLUTION
By inspection it is seen that the y
and z axes represent the principal axes of inertia since they are axes of
symmetry for the cross section. As required we have established the z
axis as the principal axis for maximum moment of inertia. The moment is
resolved into its y and z components, where
Internal Moment Components.
My =
-~(12kN · m)
Mz
3
5(12 kN · m)
=
Section Properties.
Bending Stress.
MzY
Myz
lz
ly
=
-9.60kN·m
=
7.20 kN · m
The moments of inertia about they and z axes are
c
c
c·
ly =
1 0.4 m) 0.2 m) 3
12
=
0.2667 10- 3) m4
lz =
1~ (0.2 m)(0.4 m)
=
1.067(10- 3) m4
3
Thus,
a = - -- + -7.20(103) N · m(0.2 m)
-9.60(103) N · m(-0.1 m)
- - - - -- -- - +
1.067(10- 3) m4
0.2667(10- 3) m4
ac
ao
ae
2.25 MPa
Ans.
-4.95 MPa
Ans.
=
=
7.20(103) N · m(0.2 m)
-9.60(103) N · m(0.1 m)
- - - - -- -- - +
1.067(10- 3) m4
0.2667(10- 3) m4
=
7.20(103) N · m(-0.2 m)
-9.60(1G3) N · m(0.1 m)
- - - - - -- -- - +
1.067(10- 3) m4
0.2667(10- 3) m4
=
7.20(103) N · m(-0.2 m)
-9.60(1G3) N · m(-0.1 m)
·-----~--- + -----~--4
1.067(10- 3) m
0.2667(10- 3) m4
=
=
-2.25MPa Ans.
=
4.95 MPa Ans.
The resultant normal-stress distribution has been sketched using
these values, Fig. 11- 32b. Since superposition applies, the distribution
is linear as shown.
11.5
UNSYMMETRIC BENDING
4.95 MPa
A
x
- 2.25MPa
D
(a)
(b)
Fig.11-32
M = 12kN·m
Orientation of Neutral Axis. The location z of the neutral axis (NA),
Fig. 11- 32b, can be established by proportion. Along the edge BC, we
requue
=
4.95z
"= - 79.4°
z = 0.0625 m
principal axis. According to our sign convention, 8 must be measured
from the +z axis toward the +y axis. By comparison, in Fig. 11- 32c,
8 = -tan- 1 ~ = -53.1° (or 8 = + 306.9°). Thus,
lz
tan a = -tan 8
ly
=
1.067(10- 3) m4
(
)
tan(-53.1°)
0.2667 10- 3 m4
a = -79.4°
C
B
In the same manner this is also the distance from D to the neutral axis.
We can also establish the orientation of the NA using Eq. 11- 19, which
is used to specify the angle a that the axis makes with the z or maximum
tan a
s!
~4
E ......-1-+.;D ..__
2.25 MPa
4.95 MPa
z
(0.2 m - z)
0.450 - 2.25z
A
Ans.
This result is shown in Fig.11- 32c. Using the value of z calculated above,
verify, using the geometry of the cross section, that one obtains the same
answer.
N
y
(c)
549
550
I
CHAPTER
EXAMPLE
11
BENDING
11.15
The Z-section shown in Fig.11- 33a is subjected to the bending moment of
M = 20 kN · m. The principal axes y and z are oriented as shown, such that
they represent the minimum and maximum principal moments of inertia,
Iv = 0.960(10- 3) m4 and l z = 7.54(10- 3) m4, respectively.* Determine the
normal stress at point p and the orientation of the neutral axis.
SOLUTION
z'
For use of Eq. 11- 19, it is important that the z axis represent the principal
axis for the maximum moment of inertia. (For this case most of the area
is located farthest from this axis.)
z
Internal Moment Components.
32.9° / ,
From Fig.11- 33a,
My = 20 kN · m sin 57.1° = 16.79 kN · m
Mz = 20 kN · m cos 57.1° = 10.86 kN · m
The y andl z coordinates of point P must be
determined first.Note that they' ,z' coordinates of Pare (-0.2 m, 0.35 m).
Using the colored triangles from the construction shown in Fig. 11- 33b,
we have
Bending Stress.
-300mm
(a)
yp =
-0.35 sin 32.9° - 0.2 cos 32.9°
Zp =
0.35 cos 32.9° - 0.2 sin 32.9°
=
=
-0.3580 m
0.1852 m
Applying Eq. 11- 17,
<Tp =
z'
-
MzYP
lz
My ZP
+ --
J,.
(10.86(103) N · m)(-0.3580 m)
7.54(10- 3) m4
3.76MPa
=-
z
=
N
(16.79(103) N · m)(0.1852 m)
0.960(10- ) m
+ ------~--3 4
Ans.
Using the angle 8
and the z axis, Fig.11- 33a, we have
Orientation of Neutral Axis.
7.54(10- 3) m
)
tan a = [
(
0.960 10- 3 m4
a = 85.3°
4
(b)
Fig. 11-33
y
A
=
57.1° between M
Jtan 57.1°
Ans.
The neutral axis is oriented as shown in Fig. 11- 33b.
• These values are obtained using the methods of Appendix A. (See Example A.4 or A.5.)
11. 5
UNSYMMETRIC BENDING
551
FUNDAMENTAL PROBLEMS
.11-14. Determine the bending stress developed at corners
A and 8. What is the orientation of the neutral axis?
• 11-15. Determine the maximum stress in the beam's cross
section.
y
A
r·
100 mm
y
Prob. Fll-14
Prob. Fll-15
PROBLEMS
*11-96. The member has a square cross section and is
subjected to the moment M = 850 N · m. Determine the
stress at each comer and sketch the stress distribution. Set
8 = 45°.
Prob. 11-96
11-97. The member has a square cross section and is
subjected to the moment M = 850 N · m as shown.
Determine the stress at each corner and sketch the
stress distribution. Set 8 = 300.
Prob. 11-97
552
CHAPTER
11
BE ND I NG
11-98. Consider the general case of a prismatic beam
subjected to bending-moment components M, and M , as
shown, when the x, y, z axes pass through the centroid of
the cross section. If the material is linear elastic, the
normal stress in the beam is a linear function of position
such that u = a + by + cz. Using the equilibrium
conditions 0 = JAudA, M,. = JAzudA, M, = JA - yudA ,
determine the constants a, b, and c, and show that the
normal stress can be determined from the equation
u = [- (M,I,. + M, I,,)y + (M/, + M,I,,)z]/(I,J, - I,,2),
where the moments and products of inertia are defined in
Appendix A.
11-lOL The steel shaft is subjected to the two loads. If the
journal bearings at A and B do not exert an axial force on
the shaft, determine the required diameter of the shaft if the
allowable bending stress is u allow= 180 MPa.
3if~
1.25 m
><
ym
1.25 m
Prob.11-101
11-102. The 65-mm-diameter steel shaft is subjected to the
two loads If the journal bearings at A and B do not exert an axial
force on the shaft, determine the absolute maximum bending
stress developed in the shaft.
x
Prob.11-98
11-99. Determine the bending stress at point A of the beam,
and the orientation of the neutral axis. Using the method in
Appendix A, the principal moments of inertia of the cross
section are r,. = 8.828 in4 and r,. = 2.295 in4, where z' and y'
are the principal axes. Solve the problem using Eq. 11-17.
*11-100. Determine the bending stress at point A of the
beam using the result obtained in Prob. 11-98. The moments
of inertia of the cross-sectional area about the z and y axes
are I , = I ,. = 5.561 in4 and the product of inertia of the
cross sectional area with respect to the z and y axes is
I,, = - 3.267 in4 . (See Appendix A.)
z
1.183 in.
A~ in.
~
10.s in. I
M = 3 kip· ft
1 - - -4 in.- - - 1
Probs. 11-99/100
11-103. For the section,[,. = 31.7(1o-6) m4 ,I>" = 114(10-6) m4 ,
= - 15. l(lo-6) m4 • Using the techniques outlined in
Appendix A, the member's cross-sectional area has principal
moments of inertia of I, =29.0(10-6) m4 and I,.= 117(10-6) m4 ,
calculated about the principal axes of inertia y and z,
respectively. If the section is subjected to a moment
M = 15 kN · m, determine the stress at point A using
Eq.11-17.
r,..,.
*11-104. Solve Prob.11-103 using the equation developed
in Prob. 11-98.
z'
T>-r.~H-1-f, -,__7+~c~~:~~~~~~=~--1-.118-3
Prob. 11-102
in.
z
c
60mm
1-140 nm,r t.:-::----1-:,,--1
mm60mm
Probs. 11-1031104
553
CONCEPTUAL PROBLEMS
CONCEPTUAL PROBLEMS
Cll-1. The steel saw blade passes over the drive wheel of
the band saw. Using appropriate measurements and data,
explain how to determine the bending stress in the blade.
Cll-3. Use reasonable dimensions for this hammer and a
loading to show through an analysis why this hammer failed
in the manner shown.
Prob. Cll-1
Prob. Cll-3
Cll-2. The crane boom has a noticeable taper along its
length. Explain why. To do so, assume the boom is in the
horizontal position and in the process of hoisting a load
at its end, so that the reaction on the support A becomes
zero. Use realistic dimensions and a load. to justify your
reasoning.
Cll-4. These garden shears were manufactured using an
inferior material. Using a loading of 50 lb applied normal
to the blades, and appropriate dimensions for the shears.
determine the absolute maximum bending stress in the
material and show why the failure occurred at the critical
location on the handle.
(a)
Prob. Cll-2
(b)
Prob. Cll-4
554
CHAPTER
11
BE ND I NG
CHAPTER REVIEW
Shear and moment diagrams are
graphical representations of the internal
shear and moment within a beam.
They can be constructed by sectioning
the beam an arbitrary distance x from
the left end, using the equilibrium
equations to find V and M as functions
of x, and then plotting the results. A sign
convention for positive distributed load,
shear, and moment must be followed.
Positive external distributed load
v v
---it
Positive internal shear
- -,(
M
M
Positive internal moment
Beam sign convention
It is also possible to plot the shear and
moment diagrams by realizing that
at each point the slope of the shear
diagram is equal to the intensity of the
distributed loading at the point.
Likewise, the slope of the moment
diagram is equal to the shear at the point.
dV
dx
IV = -
V = dM
dx
The area under the distributed-loading
diagram between the points represents
the change in shear.
av=
fwdx
And the area under the shear diagram
represents the change in moment.
11M =
fv dx
The shear and moment at any point
can be obtained using the method of
sections. The maximum (or minimum)
moment occurs where the shear is zero.
v
0
w = negative increasing
slope = negative increasing
I
M
CHAPTER REVIEW
A bending moment tends to produce
a linear variation of normal strain
within a straight beam. Provided the
material is homogeneous and linear
elastic, then equilibrium can be used to
relate the internal moment in the beam
to the stress distribution. The result is
the flexure formula,
y
where I and c are determined from the
neutral axis that passes through the
centroid of the cross section.
If the cross-sectional area of the beam
is not symmetric about an axis that is
perpendicular to the neutral axis, then
unsymmetrical bending will occur. The
maximum stress can be determined
from formulas, or the problem can be
solved by considering the superposition
of bending caused by the moment components My and M, about the principal
axes of inertia for the area.
)'
<T
M._y
~.z
= -- + --
I,
r,,
555
556
CHAPTER
11
BE ND I NG
REVIEW PROBLEMS
Rll-1. Determine the shape factor for the wide-flange
beam.
Rll-3. A shaft is made of a polymer having a parabolic
upper and lower cross section. If it resists a moment
of M = 125 N · m, determine the maximum bending
stress in the material (a) using the flexure formula and
(b) using integration. Sketch a three-dimensional view of
the stress distribution acting over the cross-sectional area.
Hinr: The moment of inertia is determined using Eq. A-3
of Appendix A.
y
.--!
lOOmm
I
z
Prob. Rll-1
x
Prob. Rl l-3
Rll-2. The compound beam consists of two segments
that are pinned together at B. Draw the shear and moment
diagrams if it supports the distributed loading shown.
*Rll -4. Determine the maximum bending stress in the
handle of the cable cutter at section a-a. A force of 45 lb is
applied to the handles.
45 Jb
a
w
O.SOin. H
A
0
c
B
1 -2/3 L--~1/3 L-1
Prob. Rll-2
45 Jb
Prob. Rl l-4
557
REVIEW PROBLEMS
Rl l -5. Determine the shear and moment in the beam as
functions of x. where 0 s x < 6 ft, then draw the shear and
moment diagrams for the beam.
8 kip
2 kip/ft
,.....,_so kip·ft
1
1-x-
Rll - 7. Draw the shear and moment diagrams for the shaft
if it is subjected to the vertica l loadings. The bearings at A
and B exert only vertical reactions on the shaft.
t
I
~
6 fl
B
I
-400111111--300111111- I
;I
200 111111
200 mn~
4 ft
Prob. Rll-5
150N
Prob. Rll-7
Rll~. A wooden beam has a square cross section as
shown. Determine which orientation of the beam provides
the greatest strength at resisting the moment M . What is the
difference in the resulting maximum stress in both cases?
*Rll-8. The strut has a square cross section a by a and is
subjected to the bending moment M applied at an angle 8 as
shown. Determine the maximum bending stress in terms of
a. M, and 8. What angle 8 will give the largest bending stress
in the strut? Specify the orientation of the neutral axis for
this case.
M
(b}
Prob. Rll~
Prob. Rll-8
CHAPTER
12
(©Bert Folsom/Alamy)
Railroad ties act as beams that support very large transverse shear loadings. As a
result, if they are made of wood they will tend to split at their ends, where the
shear loads are the largest.
TRANSVERSE
SHEAR
CHAPTER OBJECTIVES
•
To determine the shear stress in beams subjected to a transverse
loading.
•
To calculate the shear in fasteners used to construct beams made
from several members.
12.1
SHEAR IN STRAIGHT MEMBERS
In general, a beam will support both an internal shear and a moment. The
shear Vis the result of a transverse shear-stress distribution that acts over
the beam's cross section, Fig. 12-1. Due to the complementary property
of shear, this stress will also create corresponding longitudinal sheail" stress
that acts along the length of the beam.
Transverse
shear stress
Longitudina~shear stress ..;)~
T
Fig. 12-1
559
560
CHAPTER
12
TRANSVERSE SHEAR
ip
Boards not bonded together
(a)
Boards bonded together
(b)
Fig. 12-2
Shear connectors are " tack welded" to this
corrugated metal floor liner so that when the
concrete floor is poured, the connectors will
prevent the concrete slab from slipping on
the liner surface. The two materials will thus
act as a composite slab.
To illustrate the effect caused by the longitudinal shear stress, consider
the beam made from three boards shown in Fig. 12- 2a. If the top and
bottom surfaces of each board are smooth, and the boards are not bonded
together, then application of the load P will cause the boards to slide
relative to one another when the beam deflects. However, if the boards
are bonded together, then the longitudinal shear stress acting between
the boards will prevent their relative sliding, and consequently the beam
will act as a single unit, Fig.12- 2b.
As a result of the shear stress, shear strains will be developed and these
will tend to distort the cross section in a rather complex manner. For
example, consider the short bar in Fig.12- 3a made of a highly deformable
material and marked with horizontal and vertical grid lines. When the
shear force V is applied, it tends to deform these lines into the pattern
shown in Fig. 12- 3b. This nonuniform shear-strain distribution will cause
the cross section to warp; and as a result, when a beam is subjected to
both bending and shear, the cross section will not remain plane as
assumed in the development of the flexure formula.
v
v
~
.....
12.2 THE SHEAR FORMULA
v
v
(a) Before defom1ation
---.,... .,c; .,,.1
- \I
-
.......
~;...i,...;
~
~
r-,_\
¥I
- -'-1..J:)--"
(b) After deformation
Fig.12-3
Because the strain distribution for shear is not easily defined, as in the
case of axial load, torsion, and bending, we will obtain the shear-stress
distribution in an indirect manner. To do this we will consider the
horizontal force equilibrium of a portion of an element taken from the
beam in Fig. 12-4a. A free-body diagram of the entire element is shown
in Fig. 12-4b. The normal-stress distribution acting on it is caused by the
bending moments Mand M + dM. Here we have excluded the effects of
V, V + dv, and w(x), since these loadings are vertical and will therefore
not be involved in a horizontal force summation. Notice that ~F.r = 0 is
satisfied since the stress distribution on each side of the element forms
only a couple moment, and therefore a zero force resultant.
12.2
-
561
THE SHEAR FORMULA
IF, = 0 satisfied
...,._
dF'
d F"
Area= A '
Section plane
"
Now let's consider the shaded top por1ion of the element that has been
sectioned at y' from the neutral axis, Fig. 12-4c. It is on this sectioned
A'
plane that we want to find the shear stress. This top segment has a width t
at the section, and the two cross-sectional sides each have an area A'. The
segment's free-body diagram is shown in Fig. 12-4d. The resultant
T
moments on each side of the e lement differ by dM, so that IF,, = 0 will ~
not be satisfied unless a longitudinal shear stress Tacts over the bottom ~ \..
~
sectioned plane. To simplify the analysis, we will assume that this shear
d;r
/
,
stress is constant across the width / of the bottom face. To find the
:
:
)1,
~
~ + dM
horizontal force created by the be nding moments, we will assume that
\
:
,'
the effect of warping due to shear is small, so that it can generally be
',,~....
__ /
neglected. This assumption is particularly true for the most common case
--of a slender beam, that is, o ne that has a small depth compared to its
Three-dimensional view
length. Therefore, using the flexure formula, Eq.11- 13, we have
~ !Fx = 0:
1 .(M
{ u' dA' -
{ u ti.A - T(t dx)
) A.
) A.
0
u'
<T
~ dM)ydA' - 1.(~)y dA' -
(d~) 1.Y dA'
=
= T(t dx)
; (tdx)
=
0
I
(12-1)
Solving for T , we get
T
I
I
I
'- -
'T -
J
Profile view
ydA'
1 (dM)1
It dx
A'
-
(d)
Here V = dM / dx (Eq. 11-2). Also, the integral represents the
moment of the area A' about the neutral axis, which we will denote by
the symbol Q. Since the location of the centroid of A' is determined
from y' =
dA' /A' , we can also write
JA'y
Q
= { y dA' = y' A '
) A'
(12-2)
562
CHAPTER
12
TRANSV ERSE SHEAR
Area= A'
A
N
Fig.12-5
The final result is called the shear formula , namely
(12- 3)
With reference to Fig. 12- 5,
T
=
V
=
1=
t =
Q =
the shear stress in the member at the point located a distance y
from the neutral axis. This stress is assumed to be constant and
therefore averaged across the width t of the member
the shear force, determined from the method of sections and
the equations of equilibrium
the moment of inertia of the entire cross-sectional area calculated
about the neutral axis
the width of the member's cross section, measured at the point
where T is to be determined
y' A', where A' is the area of the top (or bottom) portion of
the member's cross section, above (or below) the section
plane where t is measured, and y' is the distance from the
neutral axis to the centroid of A '
Although for the derivation we considered only the shear stress acting
on the beam's longitudinal plane, the formula applies as well for finding
the transverse shear stress on the beam's cross section, because these
stresses are complementary and numerically equal.
12.2
Calculating Q. Of all the variables in the shear formula, Q is
usually the most difficult to define properly. Try to remember that it
represents the moment of the cross-sectional area that is above or below
the point where the shear stress is 10 be determined. It is this area A' that
is "held onto" the rest of the beam by the longitudinal shear stress as the
beam undergoes bending, Fig. 12-4d. The examples shown in Fig. 12~
will help to illustrate this point. Here the stress at point P is to be
determined, and so A' represents the dark shaded region. The value of Q
for each case is reported under each figure. These same results can also
be obtained for Q by considering A' to be the light shaded area below P,
although here y' is a negative quantity when a portion of A' is below the
neutral axis.
A
A
N
Q = j'A'
Q=y'A'
A
N
Fig. 12-6
THE SHEAR FORMULA
563
564
CHAPTER
12
TRANSVERSE SHEAR
1-b = 0.511 -j
(a)
i--- - b = 2h- - - i
7
max
VQ
= - It
(b)
Fig.12-7
Limitations on the Use of the Shear Formula. One of the
major assumptions used in the development of the shear formula is that
the shear stress is uniformly distributed over the width t at the section. In
other words, the average shear stress is calculated across the width. We
can test the accuracy of this assumption by comparing it with a more
exact mathematical analysis based on the theory of elasticity. For
example, if the beam's cross section is rectangular, the shear-stress
distribution across the neutral axis actually varies as shown in Fig. 12- 7.
The maximum value, -r' max, occurs at the sides of the cross section, and its
magnitude depends on the ratio b/h (width/depth). For sections having a
b/ h = 0.5, T 'max is only about 3% greater than the shear stress calculated
from the shear formula, Fig. 12- 7a. However, for flat sections, say
b/ h = 2, -r' max is about 40o/o greater than Tmax , Fig. 12- 7b. The error
becomes even greater as the section becomes flatter, that is, as the b/h
ratio increases. Errors of this magnitude are certainly intolerable if one
attempts to use the shear formula to determine the shear stress in the
flange of the wide-flange beam shown in Fig. 12-8.
It should also be noted that the shear formula will not give accurate
results when used to determine the shear stress at the flange- web
junction of this beam, since this is a point of sudden cross-sectional
change and therefore a stress concentration occurs here. Fortunately,
engineers must only use the shear formula to calculate the average
maximum shear stress in a beam, and for a wide-flange section this
occurs at the neutral axis, where the b/h (width/depth) ratio for the web
is very small, and therefore the calculated result is very close to the actual
maximum shear stress as explained above.
Web
Flanges
Fig.12-8
12.2
56 5
THE SHEAR FORMULA
Another important limitation on the use of the shear formula can be
illustrated with reference to Fig. 12- 9a, which shows a member having a
cross section with an irregular boundary. If we apply the shear formula to
determine the (average) shear stress T along the line AB, it will be
directed downward across this line as shown in Fig. 12- 9b. However, an
element of material taken from the boundary point B, Fig. 12- 9c, must
not have any shear stress on its outer surface. In other words, the shear
stress acting on this element must be directed tangent to the boundary,
and so the shear-stress distribution across line AB is actually directed as
shown in Fig. 12- 9d. As a result, the shear formula can only be applied at
sections shown by the blue lines in Fig. 12- 9a, because these lines
intersect the tangents to the boundary at right angles, Fig. 12- 9e.
To summarize the above points, the shear formula does not give
accurate results when applied to members having cross sections that are
short or flat, or at points where the cross section suddenly changes. Nor
should it be applied across a section that intersects the boundary of the
member at an angle other than 90°.
Stress-free
,ook"o:•~
Shear-stress distribution
from shear formula
'
(a)
'
_,.
7
(c)
(b)
Fig.12-9
(d)
566
CHAPT ER
12
TRANSV ERSE SH EAR
IMPORTANT POINTS
• Shear forces in beams cause nonlinear shear-strain distributions over the cross section, causing it to warp.
• Due to the complementary property of shear, the shear stress developed in a beam acts over the cross section
of the beam and along its longitudinal planes.
• The shear formula was derived by considering horizontal force equilibrium of a portion of a differential
segment of the beam.
• The shear formula is to be used on straight prismatic members made of homogeneous material that has
linear elastic behavior. Also, the internal resultant shear force must be directed along an axis of symmetry for
the cross section.
• The shear formula should not be used to determine the shear stress on cross sections that are short or
flat, at points of sudden cross-sectional changes, or across a section that intersects the boundary of the
member at an angle other than 90°.
PROCEDURE FOR ANALYSIS
In order to apply the shear formula, the following procedure is suggested.
Internal Shear.
• Section the member perpendicular to its axis at the point where the shear stress is to be determined, and
obtain the internal shear V at the section.
Section Properties.
• Fmd the location of the neutral axis, and determine the momen~ of inertia I of the entire cross-sectional
area about the neutral axis.
• Pass an imaginary horizontal section through the point where the shear stress is to be determined.
Measure the width t of the cross-sectional area at this section.
• The portion of the area lying either above or below this width is A' . Determine Q by using Q = y' A' .
Here )i' is the distance to the centroid of A', measured from the neutral axis. It may be helpful to realize
that A' is the portion of the member's cross-sectional area that is being "held onto the member" by the
longitudinal shear stress as the beam undergoes bending. See Figs. 12- 2 and 12-4d.
Shear Stress.
• Using a consistent set of units, substitute the data into the shear formula and calculate the shear stress T .
• It is suggested that the direction of the transverse shear stress r be established on a volume element of
material located at the point where it is calculated. This can be done by realizing that T acts on the cross
section in the same direction as V. From this, the corresponding shear stresses acting on the other three
planes of the element can then be established.
12.2
I
EXAMPLE
56 7
THE SHEAR FORMULA
12.1
The beam shown in Fig.12- lOa is made from two boards. Determine the
maximum shear stress in the glue necessary to hold the boards together
along the seam where they are joined.
SOLUTION
Internal Shear. The support reactions and the shear diagram for the
beam are shown in Fig. 12- lOb. It is seen that the maximum shear in
the beam is 19.5 kN.
The centroid and therefore the neutral axis
will be determined from the reference axis placed at the bottom of the
cross-sectional area, Fig. 12- lOa. Working in units of meters, we have
Section Properties.
(a)
26 kN
l-A
y
y =-lA
-
[0.075 mj(0.150 m)(0.030 m) + [0.165 mj(0.030 m)(0.150 m)
(0.150 m)(0.030 m) + (0.030 m)(0.150 m)
0.120 m
• - -6
m--1-l2 m~
6.5 kN
19.S kN
The moment of inertia about the neutral axis, Fig. 12- lOa, is therefore
1
] V (kN)
I = [ (0.030 m)(0.150 m)3 + (0.150 m)(0.030 m)(0.120 m - 0.075 m)2 6.5 ~--12
+ [
~
(0.150 m)(0.030 m)3 + (0.030 m)(0.150 m)(0.165 m 1
0.120 m)2]
= 27.0(10- 6 ) m4
The top board (flange) is held onto the bottom board (web) by the
glue, which is applied over the thickness t = 0.03 m. Consequently Q is
taken from the area of the top board, Fig. 12- lOa. We have
- 19.S
(b}
Q = y'A' = [0.180 m - 0.015 m - 0.120 m](0.03 m)(0.150 m)
= 0.2025(10- 3 ) m3
Shear Stress.
7:
max
Applying the shear formula ,
VQ
19.5(103 ) N(0.2025(10- 3 ) m3 )
=-=
= 488MPa
It
27.0(10- 6 ) m4(0.030 m)
·
Ans.
Plane containing glue
The shear stress acting at the top of the bottom board is shown in
Fig. 12- lOc.
NOTE: It is the glue's resistance to this longitudinal shear stress that
holds the boards from slipping at the right support.
4.88 MPa
(c)
Fig.12-10
568
I
CHAPTER
EXAMPLE
12
12.2
TRANSV ERSE S H EAR
I
Determine the distribution of the shear stress over the cross section of
the beam shown in Fig. 12- lla.
A'
(a)
(b)
SOLUTION
The distribution can be determined by finding the shear stress at an
arbitrary height y from the neutral axis, Fig. 12- 1lb, and then plotting this
function. Here, the dark colored area A' will be used for Q. * Hence
Applying the shear formula , we have
-r =
VQ
It
=
2
v(!) [ (h /4) - y2 ]b = 6V(h2
(i12bh3 )b
bh3 4
_
2)
(l)
y
This result indicates that the shear-stress distribution over the cross
section is parabolic. As shown in Fig. 12- llc, the intensity varies from
zero at the top and bottom, y = + h/2, to a maximum value at the
neutral axis, y = 0. Specifically, since the area of the cross section is
A = bh, then at y = 0 Eq. 1 becomes
Shear- stress distribution
(c)
(2)
Fig.12-11
Rectangular cross section
*The area below y can also be used [A' = b(h/2
algebraic manipulation.
+ y) ] , but doing so involves a bit more
12.2
THE SHEAR FORMULA
569
A'
~I
2 A
A
Typical shear failure of this wooden beam
occurred at the support and through the
approximate center of its cross section.
(d)
Fig. 12-11 (cont.)
This same value for Tmax can be obtained directly from the shear
formula , T = VQ/ It , by realizing that Tmax occurs where Q is largest,
since V, I, and t are constant. By inspection, Q will be a maximum
when the entire area above (or below) the neutral axis is considered;
that is, A' = bh/2 and y' = h/4, Fig.12- lld. Thus,
VQ
'Tmax
=
V(h/4)(bh/2)
[ 112bh3] b
ft =
=
V
1.5 A
By comparison, Tmax is 50% greater than the average shear stress
determined from Eq. 7-4; that is, Tavg = V/A.
It is important to realize that Tmax also acts in the longitudinal
direction of the beam, Fig. 12- lle. It is this stress that can cause a
timber beam to fail at its supports, as shown Fig. 12- llf Here
horizontal splitting of the wood starts to occur through the neutral
axis at the beam's ends, since there the vertical reactions subject the
beam to large shear stress, and wood has a low resistance to shear
along its grains, which are oriented in the longitudinal direction.
It is instructive to show that when the shear-stress distribution,
Eq. 1, is integrated over the cross section it produces the resultant
shear V. To do this, a differential strip of area dA = b dy is chosen,
Fig. 12- llc, and since r acts uniformly over this strip, we have
J
r dA
=
A
1r12 -6V (h?
-
1
- h/2
=
6~[h2 y
h
=
bh
4
3
4
)
y2 b dy
- !.;J/r/2
3
- /r/2
h) _!_(h
+h
3 8
8
6V[h2(h +
h3 4 2
2
3
3
)] =
(f)
V
570
I
CHAPTER
12
TRANSVERSE SHEAR
12.3 1
EXAMPLE
A steel wide-flange beam has the dimensions shown in Fig. 12- 12'1. If it is
subjected to a shear of V = 80 kN, plot the shear-stress distribution acting
over the beam's cross section.
20mm
-r 8 · = 1.13 MPa
B'
J,/r
[lOOf~mA
,..-------""t
I
B/ I
1---").....__..,_.,B = 22.6 MPa
I
l
c
1------1 -rc = 25.2 MPa
_______, I
I
\._
22.6MPa
1.13 MPa
(b)
(a)
SOLUTION
Since the flange and web are rectangular elements, then like the previous
example, the shear-stress distribution will be parabolic and in this case it will
vary in the manner shown in Fig. 12- 12b. Due to symmetry, only the shear
stresses at points B', B, and C have to be determined. To show how these
values are obtained, we must first determine the moment of inertia of the
cross-sectional area about the neutral axis. Working in meters, we have
I
= [
1~ (0.015 m)(0.200 m)
0.300m
._
A'
r
B
. 1·
l_l
I
I
I
y'A'
=
=
0.300 m, and A' is the dark shaded area shown in
[0.110 m](0.300 m)(0.02 m)
=
0.660(10- 3 ) m3
so that
A
Ts· =
r
s
80(103 ) N(0.660(10- 3 ) m3 )
VQs·
Its·
For point B, ts
(c)
Fig. 12-U
=
O.lOOm
N
I
QB'
t
+ (0.300 m)(0.02 m)(0.110 m) 2 ]
155.6(10-6) m4
For point B ', ts'
Fig. 12- 12c. Thus,
0.02m
]
1
(0.300 m)(0.02 m)3
12
+ 2[
=
3
=
=
0.015 m and Qs
=
1.13 MPa
Qs·, Fig.12- 12c. Hence
VQs
80(103 )N(0.660(10- 3 ) m3 )
Its
155.6(10- 6 ) m4(0.015 m)
= -- =
=
155.6(10-6) m4 (0.300 m)
=
22.6 MPa
12.2
0.02m
0.300 m
A'
I_I
-I
0.015 m
O.lOOm
I
N
A
c
(d}
Fig. 12-12 (cont.)
Note from our discussion of the limitations on the use of the shear
formula that the calculated values for both Ts· and Ts are actually very
misleading. Why?
For point C, l e = 0.015 m and A' is the dark shaded area shown in
Fig. 12- 12d. Considering this area to be composed of two rectangles,
we have
Qc
'Ly" A'
=
=
(0.110 m](0.300 m)(0.02 m)
+ l0.05 mj(0.015 m)(0.100 m)
0.735(10- 3 ) m3
=
Thus,
VQc
T
c
= T
= -- =
ltc
max
80(103 ) N(0.735(10- 3 ) m3]
155.6( 10- 6 ) m4 (0.015 m)
=
25 2 MPa
.
NOTE: From Fig. 12- 12b, the largest shear stress occurs in the web and is
almost uniform throughout its depth, varying from 22.6 MPa to 25.2 MPa.
It is for this reason that for design, some codes permit the use of
calculating the average shear stress on the cross section of the web, rather
than using the shear formula; that is,
V
80(103 ) N
=
Aw
( 0.015m ) ( 0.2m )
= -
T
avg
This will be discussed further in Chapter 15.
=
26.7MPa
THE SHEAR FORMULA
5 71
572
C HAPT ER
12
TRANSV ERSE S H EAR
PRELIMINARY PROBLEM
Pl2-1. In each case, calculate the value of Q and 1 that are
used in the shear formula for finding the shear stress at A.
Also, show how the shear stress acts on a differential volume
element located at point A.
0.3 m
y
A'
0.1 m
(d)
,...::I
1-"0.2 m
0.1 m
(a)
0.1 m
0.3 m
l
(b)
(e)
(c)
(f)
Prob. Pl2-l
12.2
THE SHEAR FORMULA
573
FUNDAMENTAL PROBLEMS
FU-1. If the beam is subjected to a shear force of
V = 100 kN, determine the shear stress at point A. Represent
the state of stress on a volume element at this point.
FU-4. If the beam is subjected to a shear force of
V = 20 kN, determine the maximum shear stress in the beam.
Prob. F12-1
FU-2. Determine the shear stress at points A and B if the
beam is subjected to a shear force of V = 600 kN. Represent
the state of stress on a volume element of these points.
Prob. F12-4
lOOmm
~
:,,+
lOOmm
!
FU-5. If the beam is made from four plates and subjected
to a shear force of V = 20 kN, determine the shear stress at
point A. Represent the state of stress on a volume element
at this point.
lOOmm
v/
50mm
Prob. F12-2
150mm
FU-3. Determine the absolute maximum shear stress in
the beam.
6 kip
150mm
3 kip
l-1ft-l-1ft-l-1ft---.B
Prob. F12-3
I
D ~·
H
.) 10.
Prob. Fl2-5
574
CHAPTER
12
TRANSVERSE SHEAR
PROBLEMS
12-L If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear stress on the web at A.
Indicate the shear-stress components on a volume element
located at this point.
12-6. The wood beam has an allowable shear stress of
'Tallow = 7 MPa. Determine the maximum shear force V that
can be applied to the cross section.
12-2. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the maximum shear stress in the
beam.
,-,-
12-3. If the wide-flange beam is subjected to a shear of
V = 20 kN, determine the shear force resisted by the web of
the beam.
50mm
SO mm
l---l-100 mm- 1 - 1
I
50mm
I,,
.l__
-,-
50mm
'"
-'- '
I
!v
200mm
1
II
IJ
Prob.12-6
B
200mm
......
20mm
Probs.12-112/3
*12-4. If the beam is subjected to a shear of V = 30 kN,
determine the web's shear stress at A and B. Indicate the
shear-stress components on a volume element located
at these points. Set w = 200 mm. Show that the neutral axis
is located at y = 0.2433 m from the bottom and
I= 0.5382(10-3) m4 .
12-7. The shaft is supported by a thrust bearing at A and a
journal bearing at B. If P = 20 kN, determine the absolute
maximum shear stress in the shaft.
*12-8. The shaft is supported by a thrust bearing at A and
a journal bearing at B. If the shaft is made from a material
having an allowable shear stress of 'Tallow = 75 MPa,
determine the maximum value for P.
12-5. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the
beam. Set w = 300 mm.
300mm
c
D
lm~--lm ~-- lm
p
p
@mm
20mm
40mm
Probs. 12-4/5
Probs.12-7/8
12.2
T HE SHEAR FORMULA
575
12-9. Determine the largest shear force V that the member
can sustain if the allowable shear stress is Ta11ow = 8 ksi.
12-13. Determine the maximum shear stress in the strut
if it is subjected to a shear force of V = 20 kN.
12-10. rr the applied shear force v = 18 kip, determine
the maximum shear stress in the member.
12-14. Determine the maximum shear force V that the
strut can support if the allowable shear stress for the
material is Tai'°"' = 40 MPa.
12mm
.J/
1
60mm
),~
!v
~'1
12mm
p..-80mm 20mm
1 in.
Probs. 12-9/10
20 mm
12-11. The overhang beam is subjected to the uniform
distributed load havi ng an intensity of w = 50 kN/m.
Determine the maximum shear stress in the beam.
dt~ l l IIEU I.l_l l l
Probs. 12-13/14
12-1.5. Sketch the intensity of the shear-stress distribution
acting over the beam's cross-sectional area, and determine the
resultant shear force acting on the segment AB. The shear force
acting at the section is V = 35 kip. Show that / NA = 872.49 in4 •
i
1-3m~l-3m~
c
.__,_ V= 35 kip
Prob. 12-11
*12-12. The beam is made from a polymer and is subjected
to a shear of V = 7 kip. Determine the maximum shear
stress in the beam and plot the shear-stress distribution over
the cross section. Report the values of the shear stress every
0.5 in. of beam depth.
Prob. 12-1.5
*12-16. Plot the shear-stress distribution over the cross
section of a rod that has a radius c. By what factor is the
maximum shear stress greater than the average shear stress
acting over the cross section?
1- 4 i n . -I
I
in.I~__,
l
I
in.-
l-
in.Ir-' V
6in.
,____,,j
Prob. 12-12
Prob. 12-16
576
CHAPTER
12
TRANSVERSE SHEAR
12-17. If the beam is subjected to a shear of V = 15 kN,
determine the web's shear stress at A and B. Indicate
the shear-stress components on a volume element
located at these points. Set w = 125 mm. Show that the
neutral axis is located at y = 0.1747 m from the bottom
and INA = 0.2182(10-3 ) m4 .
12-22. Determine the largest intensity w of the distributed
load that the member can support if the allowable shear
stress is r allow = 800 psi. The supports at A and B are
smooth.
12-23. If w = 800 lb/ft, determine the absolute maximum
shear stress in the beam. The supports at A and Bare smooth.
12-18. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the maximum shear stress in the
beam. Set w = 200 mm.
IV
12-19. If the wide-flange beam is subjected to a shear of
V = 30 kN, determine the shear force resisted by the web
of the beam. Set w = 200 mm.
t l l l ~ I l l l I l 1~ 11I }
200mm
A
-
B
3f1 - l - - - 6f1 - - - I
3ft -
Probs. 12-17/18/19
*12-20. Determine the length of the cantilevered beam so
that the maximum bending stress in the beam is equivalent
to the maximum shear stress.
Probs. 12-22/23
p
'
1 - - - - - -L - - - - -- 1
01
l-b-I
*12-24. Determine the shear stress at point Bon the web
of the cantilevered strut at section a-a.
12-25. Determine the maximum shear stress acting at
section a-a of the cantilevered strut.
Prob.12-20
12-21. If the beam is made from wood having an allowable
shear stress rauow = 400 psi, determine the maximum
magnitude of P. Set d = 4 in.
~
:J:)f_
2kN
-
250 mm - 1 - 250 mm
a
a
l -2f1 - l -2 ft - l - 2 ft -I
~J
1-1
2 in.
Prob.12-21
Probs. 12-24/25
4 kN
300 mm _ __,
12.2
12-26. Railroad ties must be designed to resist large shear
loadings. U the tie is subjected to the 34-kip rail loadings
and an assumed uniformly distributed ground reaction,
determine the intensity w for equilibrium. and calculate the
maximum shear stress in the tie at section a-a, which is
located just to the left of the rail.
34 kip
577
T HE SHEAR FORMULA
12-29. Determine the maximum shear stress in the
T-beam at the critical section where the internal shear force
is maximum.
12-30. Determine the maximum shear stress in the
T-beam at section C. Show the result on a volume element
at this point.
34 kip
a
lOkN/m
AI I I II~
-1-- - 3 ft - --1-1.s r1
A_:
l-3
D±·
1-s in.- I
0
m-l-t.s
m-~1.5
150mm
JB
m-1
1-Ll
150:~nl(Fo1m
-11- 30 nun
Prob.12-26
Probs. 12-29/30
12-27. The beam is slit longitudinally along both sides. If it
is subjected to a shear of V = 250 kN. compare the
maximum shear stress in the beam before and after the cuts
were made.
*12-28. The beam is to be cut longitudinally along both
sides as shown. U it is made from a material having an
allowable shear stress of 'Tallow = 75 MPa, determine the
maximum allowable shear force V that can be applied
before and after the cut is made.
200mm
J
12-31. The beam has a square cross section and is made of
wood having an allowable shear stress of 'Tanow = 1.4 ksi. U
it is subjected to a shear of V = 1.5 kip, determine the
smalles t dimension a of its sides.
~I.Skip
f
25mm
l
.J;s mm
I
25mm
/
25 JV2oomm
Probs.12-27128
Prob.12-31
578
CHAPTER
12
TRANSVERSE SHEAR
12. 3
Fig.12-13
SHEAR FLOW IN BUILT-UP
MEMBERS
Occasionally in engineering practice, members are "built up" from several
composite parts in order to achieve a greater resistance to loads. An
example is shown in Fig. 12- 13. If the loads cause the members to bend,
fasteners such as nails, bolts, welding material, or glue will be needed to
keep the component parts from sliding relative to one another, Fig. 12- 2.
In order to design these fasteners or determine their spacing, it is necessary
to know the shear force that they must resist. This loading, when measured
as a force per unit length of beam, is referred to as shear flow, q. *
The magnitude of the shear flow is obtained using a procedure similar
to that for finding the shear stress in a beam. To illustrate, consider finding
the shear flow along the juncture where the segment in Fig. 12- 14a is
connected to the flange of the beam. Three horizontal forces must act on
this segment, Fig.12- 14b. Two of these forces, F and F + dF, are the result
of the normal stresses caused by the moments Mand M + dM, respectively.
The third force, which for equilibrium equals dF, acts at the juncture.
Realizing that dF is the result of dM, then, like Eq. 12- 1, we have
dF = -dMJ yd.A'
I A'
F + dF
(b)
(a)
The integral represents Q, that is, the moment of the segment's area A'
about the neutral axis. Since the segment has a length dx, the shear flow,
or force per unit length along the beam, is q = dF/ dx. Hence dividing
both sides by dx and noting that V = dM/dx, Eq. 11- 2, we have
Iq
VIQ I
(12-4)
Here
q = the shear flow, measured as a force per unit length along the beam
V = the shear force, determined from the method of sections and the
equations of equilibrium
1 = the moment of inertia of the entire cross-sectional area calculated
about the neutral axis
Q
=
Y' A', where A' is the cross-sectional area of the segment that is
connected to the beam at the juncture where the shear flow is
calculated, and )i' is the distance from the neutral axis to the
centroid of A '
Fi.g .12-14
*TI1e use of the word "flow" in this terminology will become meaningful as it pertains
to the discussion in Sec. 12.4.
12.3
SHEAR FLOW IN BUILT-UP MEMBERS
579
Fastener Spacing When segments of a beam are connected by
fasteners, such as nails or bolts, their spacing s along the beam can be
determined. For example, let's say that a fastener, such as a nail, can
support a maximum shear force of F (N) before it fails, Fig. 12-lSa. If
these nails are used to construct the beam made from two boards, as
shown in Fig. 12-ISb, then the nails must resist the shear flow q (N/ m)
between the boards. In other words. the nails are used to '·hold" the top
board to the bottom board so that no slipping occurs during bending.
(See Fig. 12-2a). As shown in Fig. 12-lSc, the nail spacing is therefore
determined from
F(N) = q (N/m) s (m)
F
F
(a)
The examples that follow illustrate application of this equation.
Other examples of shaded segments connected to built-up beams by
fasteners are shown in Fig. 12-16. The shear flow here must be found at
the thick black line, and is determined by using a value of Q calculated
from A' and )i' indicated in each figure. This value of q will be resisted by
a single fastener in Fig. 12-16a, by two fasteners in Fig. 12- 16b, and by
three fasteners in Fig. 12-16c. I n other words, the fastener in Fig. 12-16a
supports the calculated value of q, and in Figs. 12- 16b and 12-16c each
fastener supports q/2 and q/3, respectively.
(b)
(c)
IMPORTANT POINT
Fig. 12-15
• Shear flow is a measure of the force per unit length along the
axis of a beam. This value is found from the shear formula and
is used to determine the shear force developed in fasteners
and glue that holds the various segments of a composite beam
together.
(a)
(b)
Fig. 12-16
(c)
580
CHAPTER
EXAMPLE
12
TRANSVERSE SHEAR
12.4
-
-
A'
( '"'B
The beam is constructed from three boards glued together as shown in
Fig. 12- 17a. If it is subjected to a shear of V = 850 kN, determine the
shear flow at B and B ' that must be resisted by the glue.
lOmm
l=f==250mm - - I
I
r
B- 1
B' -
11
I
SOLUTION
Y'B
I
I
I
N
1
.I
Section Properties. The neutral axis (centroid) will be located from
A
the bottom of the beam, Fig.12- 17a. Working in units of meters, we have
300mm
2 yA
_
'
y
V= 850kN
J
y =--=
"LA
=
J
2(0.15 m](0.3 m)(0.01 m) + [0.305 m](0.250 m)(0.01 m)
2(0.3 m)(0.01 m) + 0.250 m(0.01 m)
0.1956 m
The moment of inertia of the cross section about the neutral axis is thus
10 mm- I l -125 mm- I l -10 mm
(a)
I = 2[
+[
=
1
(0.01 m)(0.3 m)3 + (0.01 m)(0.3 m)(0.1956 m - 0.150 m)2]
12
1
(0.250 m)(0.01 m) 3 + (0.250 m)(0.01 m)(0.305 m - 0.1956 m) 2 ]
12
87.42(10- 6 ) m4
The glue at both Band B' in Fig. 12- 17a "holds" the top board to
the beam. Here
Q8
=
y8A8
=
0.2735(10- 3 ) m3
=
[0.305 m - 0.1956 m](0.250 m)(0.01 m)
Shear Flow.
I
q
I
= -
I
,,
11
c· I_
jC
N
VQ 8
A'c j
r-·Ye
A
I
=
850(103) N(0.2735(10- 3 ) m3 )
87.42( 10- 6 ) m4
=
2 66 MN/m
.
Since two seams are used to secure the board, the glue per meter
length of beam at each seam must be strong enough to resist one-half
of this shear flow. Thus,
qB = qg· =
~
=
1.33 MN/m
NOTE: If the board CC' is added to the beam, Fig. 12- 17b, then
(b)
Fig.12-17
Ans.
y and I
have to be recalculated, and the shear flow at C and C' determined
from q = V Ye A'c/ I. Finally, t!his value is divided by one-half to obtain
qcand qc .
12.3
EXAMPLE
-
SHEAR FLOW IN
BUILT-UP
5 81
MEMBERS
12.5
-
A box beam is constructed from four boards nailed together as shown
in Fig. 12- 18a. If each nail can support a maximum shear force of 30 lb,
determine the maximum spacings of the nails at Band at C so that the
beam can support the force of 80 lb.
80 lb
SOLUTION
Internal Shear.
c
If the beam is sectioned at an arbitrary point along
l - 1.Sin.
its length, the internal shear required for equilibrium is always
V = 80 lb.
,..--..j...... - ,
6 in.
Section Properties. The moment of inertia of the cross-sectional
1
area about the neutral axis can be determined by considering a
7.5 in. x 7.5 in. square minus a 4.5 in. x 4.5 in. square.
I =
~
(7.5 in.)(7.5 in.)3 - ~ (4.5 in.)(4.5 irL)3
1
1
_.., j_l.Sin.
= 229.5 in4
(a)
The shear flow at B is determined using Q8 found from the darker
shaded area shown in Fig.12- 18b. It is this "symmetric" portion of the
beam that is to be "held onto" the rest of the beam by nails on the left
side and by the fibers of the board on the right side, B '.
Thus,
Q 8 = y' A' = (3 in.](7.5 in.)(1.5 in.) = 33.75 in3
(b)
Likewise, the shear flow at C can be determined using the "symmetric"
shaded area shown in Fig. 12- 18c. We have
Qc = y'A' = (3 in.](4.5 in.)(1.5 in.) = 20.25 in3
( 1.5 in.
I
Shear Flow.
c
3 in.
N
.
VQ8
80 lb(33.75 in3 )
.
qB = - !- =
229.5 in4
= 11.76 lb1Ill.
3
qc = VQc = 80 lb(20.25 in
I
229.5 in4
)
= 7 _059 lb/in.
Fig.12-18
U;es8 = 5 in.
Ans.
And for C,
Sc =
A
(c)
These values represent the shear force per unit length of the beam
that must be resisted by the nails at Band the fibers at B', Fig.12- 18b,
and the nails at C and the fibers at C', Fig. 12- 18c, respectively. Since
in each case the shear flow is resisted at two surfaces and each nail can
resist 30 lb, for B the spacing is
30 lb
- 5.0.
Sa = (11.76/2) lb/in. - .l rn.
I
30 lb
-- 8.50.Ill. U;e sc = 8.5 in.
(7.059/2 ) lb/irL
Ans.
582
CHAPTER
12
TRANSVERSE SHEAR
EXAMPLE 12.6
-
Nails, each having a total shear strength of 40 lb, are used in a beam that can
be constructed either as in Case I or as in Case II, Fig. 12- 19. If the nails are
spaced at 9 in., determine the largest vertical shear that can be supported in
each case so that the fasteners will not fail.
0.5 in.
l_
~n'. Th~~
~~
4in. N
_I_.-~~
Case I
-1-
Fig.12-19
SOLUTION
Since the cross section is the same in both cases, the moment of inertia
about the neutral axis is calculated using one large rectangle and two
smaller side rectangles.
1
I = ~ (3 in.)(5 in.)3 - 2[ (1 in.)(4 in.)3] = 20.58 in4
12
1
Case I.
For this design a single row of nails holds the top or bottom
flange onto the web. For one of these flanges,
Q = y'A' = (2.25 in.](3 in.(0.5 in.)) = 3.375 in3
so that
40 lb
v
(3.375 in3 )
q
I ,
9 in.
20.58 in4
V = 27.1 lb
Ans.
Case II. Here a single row of nails holds one of the side boards onto
the web. Thus,
VQ
= -·
--=
y' A'
VQ
Q =
F
q =s
= -·
I '
=
(2.25 in.](1 in.(0.5 in.))
40 lb
V(l.125in3 )
--=
9 in.
20.58 in4
V = 81.3 lb
=
1.125 in3
Ans.
Or, we can also say two rows of nails hold two side boards onto the web,
so that
2(40 lb)
V(2(1.125 in3 )]
F
VQ
q = - = - ·
s
I '
9 in.
20.58 in4
v = 81.3 lb
Ans.
12.3 SHEAR
FLOW IN BUILT-UP MEMBERS
583
FUNDAMENTAL PROBLEMS
Fl2-6
The two identical boards are bolted together to
form lhc beam. Determine the maximum spacing s of the
bolts to the nearest mm if each boll has a shear strength of
15 kN.1l1e beam is subjected to a shear force of V = 50 kN.
l~
The boards are bolted together to form lhe built-up
beam. If the beam is subjected to a shear force of V = 20 kN,
determine the maxin1wu spacing s of the bolts to the
nearest mm if each bolt has a shear strength of 8 kN.
/(
LOO mm
/
7 /
JOO mm
/I
200mm
sto,~ ~
o
50mm
150
I
()
300mm
v
I/
Prob.Fil~
Prob. ;:""12-8
Two identical 20-mm-thick plates arc bolted to the
top and bottom flange to form the built-up beam. If the
beam is subjected to a shear force of V = 300 kN, determine
the maximwu spacing s of the bolts to the nearest mm if
each bolt has a shear strength of 30 kN.
r
.,_
12-1 The boards are bolted together to form the built-up
beam. If the beam is subjected to a shear force of V = 15 kip,
determine the maximwu spacings of the bolts lo the nearest
kin. if each bolt has a shear strength of 6 kip.
1 i~o.s in.
l'i
4i/(
j 3 in.
I in.
fJ'
0
0
Prob. rl..::-1
Prob. • 12-9
,.--;j in.
y
584
CHAPTER
12
TRANSVERSE SHEAR
PROBLEMS
*12-32. The beam is constructed from two boards
fastened together at the top and bottom with three rows of
nails spaced every 4 in. If each nail can support a 400-lb
shear force, determine the maximum shear force V that can
be applied to the beam.
12-33. The beam is constructed from two boards fastened
together at the top and bottom with three rows of nails spaced
every 4 in. If a shear force of V = 900 lb is applied to the
boards, determine the shear force resisted by each nail.
*12-36. The double T-beam is fabricated by welding the
three plates together as shown. Determine the shear stress
in the weld necessary to support a shear force of V = 80 kN.
12-37. The double T-beam is fabricated by welding the
three plates together as shown. If the weld can resist a shear
stress Tallow = 90 MPa, determine the maximum shear V that
can be applied to the beam.
20mm
I
--
~inX
I
!
4 in.>--
o o o
150 mm
l
v
50nm~
-
75mm
20mm
~
SO mm
-
20mm
Probs. 12-36/37
Probs. 12-32/33
12-34. The beam is constructed from three boards. If it is
subjected to a shear of V = 5 kip, determine the maximum
allowable spacing s of the nails used to hold the top and
bottom flanges to the web. Each nail can support a shear
force of 500 lb.
12-35. The beam is constructed from three boards.
Determine the maximum shear V that it can support if the
allowable shear stress for the wood is Ta1iow = 400 psi. What
is the maximum allowable spacings of the nails if each nail
can resist a shear force of 400 lb?
12-38. The beam is constructed from three boards.
Determine the maximum loads P that it can support if the
allowable shear stress for the wood is TaJJow = 400 psi. What
is the maximum allowable spacings of the nails used to hold
the top and bottom flanges to the web if each nail can resist
a shear force of 400 lb?
Al
~
p
p
.
.
C
D
I
1--6t1 - - l - 6tt - - 1 - 6t1 --I
0
1.5 in.
J-
T
Probs. 12-34/35
l lB
Prob. 12-38
12.3
12-39. A beam is constructed from three boards bolted
together as shown. Determine the shear force in each bolt if
the bolts arc spaced s = 250 mm apart and the shear is
V= 35 kN.
S HEAR FLOW IN BUILT-UP M EMBERS
585
U-42. The T-beam is constructed as shown. If each nail
can support a shear force of 950 lb, determine the maximum
shear force V that the beam can support and the
corresponding maximum nail spacings to the nearest k in.
The allowable shear stress for the wood is Ta1io.. = 450 psi.
A
25m/ ,,l_
25mm 1m
i:J~
'1
350mm
~
~~
2 in.
Prob. 12-39
Prob. 12-42
*12-40. The si mply supported beam is built up from three
boards by nailing them together as shown. The wood has an
allowable shear stress of Tanow = 1.S MPa, and an allowable
bending stress of O'auow = 9 MPa The nails are spaced at
s = 75 mm, and each has a shear strength of 1.5 kN.
Determine the maximum allowable force P that can be
applied to the beam.
12-4L The simply supported beam is built up from three
boards by nailing them together as shown. If P = 12 kN,
determine the maximum allowable spacing s of the nails to
support that load. if each nail can resist a shear force of LS kN.
U-43. The box beam is constructed from four boards that
are fastened together using nails spaced along the beam
every 2 in. If each nail can resist a shear force of SO lb,
determine the largest force P that can be applied to the
beam without causing failure of the nails.
*12-44. The box beam is constructed from four boards
that are fastened together using nails spaced along the
beam every 2 in. If a force P = 2 kip is applied to the beam,
determine the shear force resisted by each nail at A and B.
p
11
A
0000000000000
B
- - - 1 m - - - - -- 1 m--
." l==f25 mm
I
~mm-~
'
--l
•
}"'
·I
1 25 mm
Probs. 12-40/41
Probs. 12-43/44
586
CHAPTER
12
TRANSVERSE SHEAR
12-45. The member consists of two plastic channel strips
0.5 in. thick, glued together at A and B. If the distributed
load has a maximum intensity of w0 = 3 kip/ft, determine
the maximum shear stress resisted by the glue.
12-47. The beam is made from four boards nailed together
as shown. U the nails can each support a shear force of
100 lb., determine their required spacing s' and s if the
beam is subjected to a shear of V = 700 lb.
- - -6ft-------l1---6ft- - -
10 in.
l
,.--(
3 in.
,.....I l/l.Sin.
~
3 in.
>
Prob.12-45
Prob.12-47
*12-48. The beam is made from three polystyrene strips
that are glued together as shown. If the glue has a shear
strength of 80 kPa, determine the maximum load P that can
be applied without causing the glue to Jose its bond.
30mm
12-46. The member consists of two plastic channel strips
0.5 in. thick, glued together at A and B. If the glue can
support an allowable shear stress of Taiiow = 600 psi,
determine the maximum intensity w0 of the triangular
distributed loading that can be applied to the member
based on the strength of the glue.
p
-,40mm
!p
I
-I
- 20mm
60mm
I-,
40mm
~0.8m -l- 1 m ~- 1 m -~ 0.8m~
_I_
Prob.12-48
12-49. The timberT-beam is subjected to a load consisting
of n concentrated forces, P,,. If the allowable shear V,,a;1 for
each of the nails is known , write a computer program that
will specify the nail spacing between each load. Show an
application of the program using the values L = 15 ft,
a 1 = 4 ft, Pt = 600 lb, a 2 = 8 ft, Pi = 1500 lb, b1 = 1.5 in.,
h.1 = 10 in, bi = 8 in, h.2 = 1 in, and V,,.; 1 = 200 lb.
,.--(
l
3 in.
~
3 in.
>
Prob.12-46
P2
1- sr1
P,.
l Al
--------------<~B
A.,.,
i
'f ., -l~
Prob.12-49
i
587
CHAPTER REVIEW
CHAPTER REVIEW
Transverse shear stress in beams is determined indirectly by
using the flexure formula and the relationship between
moment and shear (V = dM/dx). The result is the shear
formula
'T
Area= A '
VQ
= Ir
A
In particular, the value for Q is the moment of the area A'
about the neutral axis, Q )i' A '. This area is the portion of
=
N
the cross-sectional area that is .. held onto" the beam above
(or below) the thickness r where 'T is to be determined.
If the beam has a rectangular cross section, then the
shear-stress distribution will be parabolic, having a maximum
value at the neutral axis. For this special case, the maximum
shear stress can be determined using
'Tmox
v
=15• A
Shear-stress distribution
Fasteners, such as nails, bolts, glue, or weld, are used to
connect the composite parts of a .. built-up" section. The
shear force resisted by these fasteners is determined from
the shear Oow, q, or force per unit length, that must be
supported by th e bcam. l11c shear flow is
VQ
q=1
A
588
CHAPT ER
12
TRANSV ERSE SHEAR
REVIEW PROBLEMS
R12-L The beam is fabricated from four boards nailed
together as shown. Determine the shear force each nail
along the sides C and the top D must resist if the nails are
uniformly spaced at s = 3 in. The beam is subjected to a
shear of V = 4.5 kip.
R12-3. The member is subjected to a shear force of
V = 2 kN. Determine the shear flow at points A , B, and C.
The thickness of each thin-walled segment is 15 mm.
-,-
B
lOOmm
_I _
300mm
12 in.
1
V= 2kN
Prob. R12-3
,,-! !-" 1 in .
Prob. R12-1
*R12-4. The beam is constructed from four boards glued
together at their seams. If the glue can withstand 75 lb/in.,
what is the maximum vertical shear V that the beam can
support? What is the maximum vertical shear V that the
beam can support if it is rotated 90° from the position shown?
R12-2. The T-beam is subjected to a shear of V = 150 kN.
Determine the amount of this force that is supported by
the webB.
=
-,
3 in.
•
1
f'-'-':'l-":......,i;::-1- ,
-
V= 150 kN
B
- j - 4in. - f ( 0.5 in.
0.5 in.
Prob. R12-2
o.5 in.
~
__.,
Prob. R12-4
REVIEW PROBLEMS
R12-S. Determine the shear stress at points B and C on
the web of the beam located at section a-a.
R12-6. Determine the maximum shear stress acting at
section a-a in the beam.
589
*RU-3. The member consists of two triangular plastic
strips bonded together along AB. If the glue can support an
allowable shear stress of 'T.iiow = 600 psi, determine the
maximum vertical shear V that can be applied to the
member based on the strength of the glue.
80001b
150 lb/ft
!!
r
A _u_
I
C •
a
4 ft
/01
jo
1.5 fl
'
I
4 fl
1.5 fl
21inl.
D
0.75 in.
-=- lJ
c£
0.5 in.jJcB
6 in.
I 6lin.
1
I
r=1-r1in.
4 m.
Prob. R1 2-8
0.75
Probs. R12-5/6
R12-9. U the pipe is subjected to a shear of V = 15 kip.
determine the maximum shear stress in the pipe.
R12-7. The beam supports a vertical shear of V = 7 kip.
Determine the resultant force this develops in segment AB
of the beam.
0.5 in.
0.5 in.
2.3 in.
V=7 kip
Prob. R12-7
v
Prob. R1 2-9
CHAPTER
13
(© lmageBroker/Alamy)
The offset hanger supporting this ski gondola is subjected to the combined
loadings of axial force and bending moment.
COMBINED
LOADINGS
CHAPTER OBJECTIVES
•
To analyze the stresses in thin-walled pressure vessels.
•
To show how to find the stresses in members subjected to combined
loadings.
13.1
THIN-WALLED PRESSURE
VESSELS
Cylindrical or spherical pressure vessels are commonly used in industry
to serve as boilers or storage tanks. The stresses acting in the wall of these
vessels can be analyzed in a simple manner provided it has a thin wall, that
is, the inner-radius-to-wall-thickness ratio is 10 or more (r/t > 10).
Specifically, when r/t = 10 the results of a thin-wall analysis will predict
a stress that is approximately 4% less than the actual maximum stress in
the vessel. For larger r/ t ratios this error will be even smaller.
In the following analysis, we will assume the gas pressure in the vessel
is the gage pressure, that is, it is the pressure above atmospheric pressure,
since atmospheric pressure is assumed to exist both inside and outside
the vessel's wall before the vessel is pressurized.
Cylindrical pressure vessels, such as this gas
tank, have semispherical end caps rather
than Oat ones in order to reduce the stress
in the tank.
591
592
CHAPTER
13
COMBINED LOAD I NGS
Cylindrical Vessels. The cylindrical vessel in Fig. 13- la has a wall
)'
thickness t, inner radius r, and is subjected to an internal gas pressure p.
To find the circumferential or hoop stress, we can section the vessel by
planes a, b, and c. A free-bodly diagram of the back segment along with
its contained gas is then shown in Fig. 13- lb. Here only the loadings in
the x direction are shown. They are caused by the uniform hoop stress u 1,
acting on the vessel's wall, and the pressure acting on the vertical face of
the gas. For equilibrium in the x direction, we require
2F.x
(a)
=
O·,
2(u1(t dy)] - p(2r dy)
= 0
(13- 1)
The longitudinal stress can be determined by considering the left
portion of section b, Fig. 13- la. As shown on its free-body diagram,
Fig. 13- lc, u 2 acts uniformly throughout the wall, and p acts on the
section of the contained gas. Since the mean radius is approximately
equal to the vessel's inner radius, equilibrium in they direction requires
(b)
2F.y
=
O·,
I
(13-2)
For these two equations,
<Ti,
(c)
Fig.13-1
u 2 = the normal stress in the hoop and longitudinal directions,
respectively. Each is assumed to be constant throughout the
wall of the cylinder, and each subjects the material to tension.
p = the internal gage pressure developed by the contained gas
r = the inner radius of the cylinder
t = the thickness of the wall (r /t =::: 10)
13.1
THIN-WALLED PRESSURE VESSELS
59 3
By comparison, note that the hoop or circumferential stress is twice as
large as the longitudinal or axial stress. Consequently, when fabricating
cylindrical pressure vessels from rolled-formed plates, it is important that
the longitudinal joints be designed to carry twice as much stress as the
circumferential joints.
Spherical Vessels. We can analyze a spherical pressure vessel in a
similar manner. If the vessel in Fig. 13-2a is sectioned in half, the resulting
free-body diagram is shown in Fig. 13-2b. Like the cylinder, equilibrium
in they direction requires
This thin-walled pipe was subjected to an
excessive gas pressure that caused it to rupture
in the circumferential or hoop direction. The
stress in this direction is twice that in the axial
direction as noted by Eqs. 13-1 and 13-2.
(13- 3)
This is the same result as that obtained for the longitudinal stress in the
cylindrical pressure vessel, although this stress will be the same regardless
of the orientation of the hemispheric free-body diagram.
y
x
Limitations. The above analysis indicates that an element of material
taken from either a cylindrical or a spherical pressure vessel is subjected
to biaxial stress, i.e., normal stress existing in only two directions.
Actually, however, the pressure also subjects the material to a radial
stress, u3 , which acts along a radial line. This stress has a maximum value
equal to the pressure p at the interior wall and it decreases through the
wall to zero at the exterior surface of the vessel, since the pressure there
is zero. For thin-walled vessels, however, we will ignore this stress
component, since our limiting assumption of r/t = 10 results in a 2 and
u 1 being, respectively, 5 and 10 times higher than the maximum radial
stress, (u3)max = p . Finally, note that if the vessel is subjected to an
external pressure, the resulting compressive stresses within the wall may
cause the wall to suddenly collapse inward or buckle rather than causing
the material to fracture.
(a)
I
(b)
Fig.13-2
594
I
CHAPTER
EXAMPLE
13
COMBINED LOAD I NGS
13.1
A cylindrical pressure vessel has an inner diameter of 4 ft and a thickness
of ~ in. Determine the maximum internal pressure it can sustain so that
neither its circumferential nor its longitudinal stress component exceeds
20 ksi. Under the same conditions, what is the maximum internal pressure
that a similar-size spherical vessel can sustain?
SOLUTION
The maximum stress occurs in the
circumferential direction. From Eq.13-1 we have
Cylindrical Pressure Vessel.
<T I -
pr
-
·
t '
. ;·
20 kIp Ill2
=
p(24 in.)
I .
210.
p = 417psi
Ans.
Note that when this pressure is reached, from Eq. 13-2, the stress in the
longitudinal direction will be u 2 = (20 ksi ) = 10 ksi, and the maximum
stress in the radilll direction is at the inner wall of the vessel,
(u3)max = p = 417 psi.Thisvalueis48tirnessmallerthanthecircurnferential
stress (20 ksi), and as stated earlier, its effect will be neglected.
!
Here the maximum stress occurs in any two
perpendicular directions on an element of the vessel, Fig. 13- 2a. From
Eq. 13- 3, we have
Spherical Vessel.
<T2 -
-
pr
- ·
2t'
.
.
2
20k1p/m
p(24 in.)
=
(
1 .
2 2m.
p = 833 psi
)
Ans.
NOTE: Although it is more difficult to fabricate, the spherical pressure
vessel will carry twice as much internal pressure as a cylindrical vessel.
13.1
595
THIN-WALLED PRESSURE VESSELS
PROBLEMS
13-L A spherical gas tank has an inner radius of r = 1.5 m.
If it is subjected to an internal pressure of p = 300 kPa,
determine its required thickness if the maximum normal stress
is not to exceed 12 MPa.
13-5. Air pressure in the cylinder is increased by exerting
forces P = 2 kN on the two pistons, each having a radius
of 45 mm. If the cylinder has a wall thickness of 2 mm,
determine the state of stress in the wall of the cylinder.
13-2. A pressurized spherical tank is made of 0.5-in.-thick
steel. If it is subjected to an internal pressure of p = 200 psi,
determine its outer radius if the maximum normal stress is
not to exceed 15 ksi.
13-6. Determine the maximum force P that can be exerted
on each of the two pistons so that the circumferential stress in
the cylinder does not exceed 3 MPa. Each piston has a radius
of 45 mm and the cylinder has a wall thickness of2 mm.
13-3. The thin-walled cylinder can be supported in one of
two ways as shown. Determine the state of stress in the wall
of the cylinder for both cases if the piston P causes the
internal pressure to be 65 psi. The wall has a thickness of
0.25 in., and the inner diameter of the cylinder is 8 in.
'
p
- -P
1-----'"I
- 8 in. -
p
8 in.
Probs. 13-5/6
.·
(a)
(b)
Prob.13-3
*13-4. The tank of the air compressor is subjected to
an internal pressure of 90 psi. If the inner diameter of the
tank is 22 in., and the wall thickness is 0.25 in., determine
the stress components acting at point A. Draw a volume
element of the material at this point, and show the results
on the element.
13-7. A boiler is constructed of 8-mm-thick steel plates
that are fastened together at their ends using a butt joint
consisting of two 8-mm cover plates and rivets having a
diameter of 10 mm and spaced 50 mm apart as shown. If the
steam pressure in the boiler is 1.35 MPa, determine (a) the
circumferential stress in the boiler's plate away from the seam,
(b) the circumferential stress in the outer cover plate along
the rivet line a-a, and (c) the shear stress in the rivets.
a
Prob.13-4
Prob.13-7
596
CHAPTER
13
CO MBIN ED LOADINGS
*13-8. The steel water pipe has an inner diameter of 12 in.
and a wall thickness of 0.25 in. If the valve A is opened and
the flowing water has a pressure of 250 psi as it passes point B,
determine the longitudinal and hoop stress developed in the
wall of the pipe at point B.
13-9. The steel water pipe has an inner diameter of 12 in.
and a wall thickness of 0.25 in. If the valve A is closed and
the water pressure is 300 psi, determine the longitudinal and
hoop stress developed in the wall of the pipe at point B.
Draw the state of stress on a volume element located on
the wall.
13-lL The gas pipe line is supported every 20 ft by
concrete piers and also lays on the ground. If there are rigid
retainers at the piers that hold the pipe fixed, determine the
longitudinal and hoop stress in the pipe if the temperature
rises 60° F from the temperature at which it was installed.
The gas witihin the pipe is at a pressure of 600 lb/ in2 • The
pipe has an inner diameter of 20 in. and thickness of 0.25 in.
The material is A-36 steel.
i - - - - - -20 ft - - - - - -1
1
~B
A
~~~1'-~~~~~~--<
li~~,
Prob.13-11
Probs. 13-819
13-10. The A-36-steel band is 2 in. wide and is secured
around the smooth rigid cylinder. If the bolts are tightened
so that the tension in them is 400 lb, determine the normal
stress in the band, the pressure exerted on the cylinder, and
the distance half the band stretches.
*13-12. A pressure-vessel head is fabricated by welding
the circular plate to the end of the vessel as shown. If the
vessel sustains an internal pressure of 450 kPa, determine
the average shear stress in the weld and the state of stress in
the wall oft he vessel.
l -450mm -I
f 10mm
/
8 in.
Prob.13-10
20mm
Prob.13-12
1 3.1
13-13. An A-36-steel hoop has an inner diameter of
23.99 in., thickness of 0.25 in., and width of 1 in. If it and the
24-in.-diameter rigid cylinder have a temperature of 65° F,
determine the temperature to which the hoop should be
heated in order for it to just slip over the cylinder. What is the
pressure the hoop exerts on the cylinder, and the tensile stress
in the ring when it cools back down to 65° F?
THIN-WALLED PRESSURE V ESSELS
597
*13-16- Two hemispheres having an inner radius of 2 ft
and wall thickness of 0.25 in. are fitted together, and the
inside pressure is reduced to - 10 psi. If the coefficient
of static friction is µ,, = 0.5 between the hemispheres,
determine (a) the torque T needed to initiate the rotation
of the top hemisphere relative to the bottom one, (b) the
vertical force needed to pull the top hemisphere off the
bottom one, and (c) the horizontal force needed to slide
the top hemisphere off the bottom one.
Prob.13-13
13-14. The ring, having the dimensions shown, is placed
over a flexible membrane which is pumped up with a
pressure p. Determine the change in the inner radius of the
ring after this pressure is applied. The modulus of elasticity
for the ring is E.
Prob.13-16
w
p
Prob.13-14
13-15. The inner ring A has an inner radius r 1 and outer
radius r 2 • The outer ring B has an inner radius r3 and an outer
radius r 4 , and r 2 > r3 . If the outer ring is heated and then
fitted over the inner ring, determine the pressure between
the two rings when ring B reaches the temperature of the
inner ring. The material has a modulus of elasticity of E and
a coefficient of thermal expansion of a.
A
13-17. In order to increase the strength of the pressure vessel,
filament winding of the same material is wrapped around the
circumference of the vessel as shown. If the pretension in the
filament is T and the vessel issubjected to an internal pressurep,
determine the hoop stresses in the filament and in the wall
of the vessel. Use the free-body diagram shown, and assume
the filament winding has a thickness 1' and width w for a
corresponding length L of the vessel.
B
Prob.13-15
Prob.13-17
598
CHAPTER
13
COMBINED LOAD I NGS
13.2 STATE OF STRESS CAUSED BY
COMBINED LOADINGS
In the previous chapters we showed how to determine the stress in a
member subjected to either an internal axial force, a shear force, a bending
moment, or a torsional moment. Most often, however, the cross section of
a member will be subjected to several of these loadings simultaneously,
and when this occurs, then the method of superposition should be used to
determine the resultant stress. The following procedure for analysis
provides a method for doing this.
PROCEDURE FOR ANALYSIS
This chimney is subjected to the combined
internal loading caused by the wind and the
chimney's weight.
Here it is required that the material be homogeneous and behave in
a linear elastic manner. Also, Saint-Venant's principle requires that
the stress be determined at a point far removed from any
discontinuities in the cross section or points of applied load.
Internal Loading.
• Section the member perpendicular to its axis at the point
where the stress is to be determined; and use the equations of
equilibrium to obtain the resultant internal normal and shear
force components, and the bending and torsional moment
components.
• The force components should act through the centroid of the
cross section, and the moment components should be calculated
about centroidal axes, which represent the principal axes of
inertia for the cross section.
Stress Components.
• Determine the stress component associated with each internal
loading.
Normal Force.
• The normal force is related to a uniform normal-stress
distribution determined from <T = N /A.
13.2
STATE OF S TRESS CAUSED BY COMBINED LOADINGS
599
Shear Force.
• The shear force is related to a shear-stress distribution
determined from the shear formula , T = V Q / It.
Bending Moment.
• For straight m embers the bending moment is related to a
normal-stress distribution that varies linearly from zero at
the neutral axis to a maximum at the outer boundary of the
member. This stress distribution is determined from the
flexure formula,u = -My/ I. If the member is curved, the
stress distribution is nonlinear and is determined from
u = My/ (A e(R - y)].
Torsional Moment.
• For circular shafts and tubes the torsional moment is
related to a shear-stress distribution that varies linearly
from zero at the center of the shaft to a maximum at the
shaft's outer boundary. This stress distribution is
determined from the torsion formula,T = Tp/ l.
Thin-Walled Pressure Vessels.
• If the vessel is a thin-walled cylinder, the internal pressure
p will cause a biaxial state of stress in the material such
that the hoop or circumferential stress component is
u 1 = pr/t, and the longitudinal stress component is
u 2 = pr/21. If the vessel is a thin-walled sphere, then the
biaxial state of stress is represented by two equivalent
components, each having a magnitude of u 2 = pr/21.
Superposition.
• Once the normal and shear stress components for each loading
have been calcuJated , use the principle of superposition and
determine the resultant normal and shear stress components.
• Represent the results on an element of material located at a
point, or show the results as a distribution of stress acting over
the member's cross-sectional area.
Problems in this section, which involve combined loadings, serve as a
basic review of the application of the stress equations mentioned above.
A thorough understanding of how these equations are applied, as
indicated in the previous chapters, is necessary if one is to successfully
solve the problems at the end of this section. The following examples
should be carefully studied before proceeding to solve the problems.
When a pretension force F is developed in
the blade of this coping saw, it wiU produce
both a compressive force F and bending
moment M at 1he section AB o( the frame.
The ma1erial must therefore resist the
normal stress produced by both of these
loadings.
600
I
CHAPTER
EXAMPLE
13
COMBINED LOADINGS
13.2
I
150Jb
I 5 in. I 5 in. I
--:7----::7.:::::....---..i:~:f;~2
in.
2 in.
A force of 150 lb is applied to the edge of the member shown in Fig. 13- 3a.
Neglect the weight of the member and determine the state of stress at
points B and C.
SOLUTION
The member is sectioned through B and C,
Fig. 13- 3b. For equilibrium at the section there must be an axial force of
150 lb acting through the centroid and a bending moment of 750 lb · in.
about the centroidal principal axis, Fig. 13-3b.
Internal Loadings.
c
B
Stress Components.
The uniform normal-stress distribution due to the
normal force is shown in Fig. 13- 3c. Here
N
150 lb
.
CT = - = ( O. )( . ) = 3.75 psi
A
1 m. 4m.
Normal Force.
(a)
Fig.13-3
The normal-stress distribution due to the bending
moment is shown in Fig. 13- 3d. The maximum stress is
Bending Moment.
150Jb
Mc
750 lb · in. (5 in.)
.
<Tmax = = 1
=
11.25
psi
3
I
12 (4in.)(10in.)
Superposition.
c
B
750 lb·in.
1501b
(b)
Algebraically adding the stresses at B and C, we get
.
.
.
N
Mc
<Ts = - A + I
= - 3.75 psi + 11.25 psi = 7.5 psi
(
. )
tension
Ans.
.
.
.(
. )
Mc
<Tc = - N
A - I= - 3.75 psi - 11.25 psi = -15 psi compress10n
Ans.
These results are shown in Figs. 13- 3/ and 13- 3g.
NOTE: The resultant stress distribution over the cross section is shown in
Fig. 13- 3e, where the location of the line of zero stress can be determined
by proportional triangles; i.e.,
7.5 psi
15 psi
.
x
- (lOin. -x); x = 3.33m.
~A±l:b :r.Jc
+
c
7.5p~I- xJM
3.75 psi
!1 15psi
-I
(10 in. - x)
Normal force
Bending moment
Combined loading
(c)
(d)
(e)
B~i c ~i
7.5 psi
(f)
15 psi
(g)
13.2
-
EXAMPLE
STATE OF STRESS CAUSED BY COMBINED LOADINGS
13.3
-
The gas tank in Fig. 13--4a has an inner radius of 24 in. and a thickness of
0.5 in. If it supports the 1500-lb load at its top, and the gas pressure within it
is 2 lb /in2, determine the state of stress at point A.
SOLUTION
Internal Loadings. The free-body diagram of the section of the tank
above point A is shown in Fig. 13--4b.
Stress Components.
Since r /t = 24 in./0.5 in. = 48 > 10, the tank
is a thin-walled vessel.Applying Eq.13- 1, using the inner radius r = 24 in.,
we have
Circumferential Stress.
pr
u1
=- =
l
2 lb/in2 (24 in.)
0.5 in.
=
.
96 psi
Ans.
Here the wall of the tank uniformly supports the
load of 1500 lb (compression) and the pressure stress (tensile). Thus, we
have
Longitudinal Stress.
N
pr
A
2t
~ = - - + -= -
=
1500 lb
1T[(24.5 in.) 2 - (24 in.)2]
2 lb/in2 (24 in.)
+
2 (0.5 in.)
28.3 psi
Ans.
Point A is therefore subjected to the biaxial stress shown in Fig. 13--4c.
1500lb
I=
1500lb
0.5 in.
_aIJJJ1,,
A
A
!28.3 psi
____r=-,
~­
LJJJ 96 psi
Al
(a)
(b)
Fig. 13-4
(c)
601
602
I
CHAPTER
EXAMPLE
13
COMBINED LOAD I NGS
13.4 1
The member shown in Fig. 13-5a has a rectangular cross section. Determine
the state of stress that the loading produces at point C and point D.
c
D
1 -1.5m l
~mm
~ct
A
12r mm
j
c
125mm
1.5 m
l
50mm
1 - - - - -4 m - - - - - - · 1 - -
(a)
1-----4
m _ _ _ _ __,
16.45 kN
1
0.75m
21.93 kN
0.75m
J
(b}
lm
1-i. m~I v
5
ji'M
16.45kN1
N
21.93 kN
(c)
Fig.13-5
SOLUTION
The support reactions on the member have been
determined and are shown in Fig. 13- 5b. (As a review of statics, apply
2MA = 0 to show F8 = 97.59 kN.) If the left segment AC of the member
is considered, Fig.13- 5c, then the resultant internal loadings at the section
consist of a normal force, a shear force, and a bending moment. They are
Internal Loadings.
N
=
16.45 kN
V
=
21.93 kN
M
=
32.89 kN · m
13.2
STATE OF S TRESS CAUSED BY C O MBINED LOADINGS
c
+
+
Normal force
Shear force
Bending moment
(d)
(e)
(f)
Fig. 13-5 (cont.)
Stress Components at C.
The unifo rm normal-stress distribution acting ove r the
cross section is produced by the normal force, Fig. 13- 5d. At point C:
Normal Force.
N
<Tc = -A
16.45(103) N
= -(0-.0-5_0_m_)_(0-.25
- 0_m_)
= 1.32 MPa
Here the area A' = 0, since point C is located at the top
of the me mber. Thus Q = y A ' = 0, Fig.13-5e. The shear stress is therefore
Shear Force.
Tc = 0
Point C is locate d at y = c
ne utral axis, so the bending stress at C, Fig. 13-5/, is
Bending Moment.
Mc
u: = c
I
=
=
0.125 m from the
(32.89(103) N · m)(0.125 m)
[
• = 63.16MPa
fi (0.050 m) (0.250 m)')
Superposition. There is no shear-stress co mponent.Adding the normal
stresses gives a compressive stress at C having a value of
<rc
= 1.32 MPa
+ 63.16 MPa
=
64.5 MPa
Ans.
This result, acting on an e lement at C, is shown in Fig. 13- 5g.
64.5 MPa
(g)
Stress Components at D.
Normal Force.
This is the same as at C, <r 0
=
1.32 MPa, Fig. 13-5d.
Since D is at the neutral axis, and the cross section is
re ctangular, we can use the special form of the shear formula, Fig. 13-5e.
Shear Force.
V
To = 1.5 A
Bending Moment.
=
21.93(103) N
1.5 (0. 25 m)(0.05 m)
= 2.63 MPa
Ans.
He re D is on the neutral axis and so er0 = 0.
Superposition. The resultant stress on the element is shown in Fig. 13-5h.
2.63 MPa
--\ j l - t .32 MPa
(h)
603
604
I
CHAPT ER
EXAMPLE
13
13.s
COMBIN ED LOAD I NGS
I
The solid rod shown in Fig. 13--6a has a radius of 0.75 in. If it is subjected to
the force of 500 lb, determine the state of stress at point A .
SOLUTION
The rod is sectioned through point A. Using the
free-body diagram of segment AB, Fig. 13- 6b, the resultant internal
loadings are determined from the equations of equilibrium.
Internal Loadings.
IF). = O;
'I.Mz
=
500 lb - Ny = O; Ny = 500 lb
O; 500 lb(14 in.) - Mz
Mz
O;
=
=
7000 lb · in.
In order to better "visualize" the stress distributions due to these loadings,
we can consider the equal but opposite resultants acting on segment AC,
Fig. 13--6c.
Stress Components.
Normal Force. The normal-stress distribution is shown in Fig. 13--6d.
For point A, we have
Ny = 500 lb
(aA)v
·
=
N
A
=
(
'TT
500 1
.b )2
0.75 m.
=
283 psi
=
0.283 ksi
Bending Moment. For the moment, c = 0.75 in., so the bending stress
at point A , Fig. 13--6e, is
Mc
(aA) y =
(b)
=
I=
7000 lb · in.(0.75 in.)
[!7r(0.75 in.)4 )
21126 psi
=
21.13 ksi
Superposition. When the above results are superimposed, it is seen that
an element at A, Fig. 13- 6/, is subjected to the normal stress
(aA)y = 0.283 ksi + 21.13 ksi
=
Ans.
21.4 ksi
7000 lb·in.
~
~500lb
(c)
+
~1.13 ksi ~1.4ksi
Normal force
Bending moment
(d)
(e)
Fig.13-6
(f)
13.2
EXAMPLE
STATE OF S TRESS CAUSED BY COMBINED LOADINGS
13.6
The solid rod shown in Fig. 13-7a has a radius of 0.75 in. If it is subjected to
the force of 800 lb, determine the state of stress at point A.
SOLUTION
Internal Loadings. The rod is sectioned through point A. Using the
free-body diagram of segment AB, Fig. 13-7b, the resultant internal
loadings are determined from the equations of equilibrium. Take a
moment to verify these results. The equal but opposite resultants are shown
acting on segment AC, Fig. 13-7c.
"i.£. = O; Vz - 800 lb = 0; Vz = 800 lb
"i.Mx = O; Mx - 800 lb(lO in.) = O; Mr = 8000 lb· in.
"i.M,, = O; -M,, + 800 1b( 14in.) = O; M,, = 112001b·in.
Stress Components.
Shear Force. The shear-stress distribution is shown in Fig. 13-7d. For
point A , Q is determined from the gray shaded semicircular area. Using
the table in Appendix B, we have
Q = J' A' = 4(0.757T in.) [ 1 7r(0.75 in.)-'] = 0.28125 in-3
3
2
so that
VQ
800 lb(0.28125 in3)
(Ty:)A
=fr=
= 604 psi
c
/
(a)
M1
~
11 200 lb·in .
M,
~
8000 lb·inp .
[ ~(0.75in.) 4 ]2(0.75in.)
= 0.604 ksi
~~
x,./"'
Bending Moment. Since point A lies on the neutral axis, Fig.13-7e, the
bending stress is
<TA = 0
(b)
Torque. At point A,pA = c = 0.75 in.,Fig.13-7fThus the shear stress is
Tc
11 200 lb· in.(0.75 in.)
.
.
(r,.,)A = - =
['
. 4] = 16901 psi= l6.90ks1
J
27T(0.75 m.)
Fig. 13-7
Superposition. Here the element of material at A is subjected only to
a shear stress component, Fig. 13-7g, where
(r,.z)A = 0.604 ksi + 16.90 ksi = 17.5 ksi
Ans.
+
+
800 lb
11 200 lb· in.
17.5 ksi
16.90 ksi
(c)
Shear force
Bending moment
Torsional moment
(d)
(e)
(f)
(g)
605
606
CHAPTER
13
COMBINED LOADINGS
PRELIMINARY PROBLEMS
P13-L In each case, determine the internal loadings that
act on the indicated section. Show the results on the left
segment.
200N
SOON
2m
(d)
Prob. P13-1
P13-2. The internal loadings act on the section. Show the
stress that each of these loads produce on differential
elements located at point A and point B.
(a)
v
lOON
300N
(b)
N
(a)
M
(c)
(b)
Prob. P13-2
13.2
STATE OF S TRESS CAUSED BY COMBINED LOADINGS
607
FUNDAMENTAL PROBLEMS
.13-1. Determine the normal stress at corners A and B of
the column.
• 13-3. Determine the state of stress at point A on the
cross section of the beam at section a-a. Show the results in
a differential element at the point.
500kN
300kN
B
m~ Nm
Prob. F13-1
Pr
• 13-2 Determine the state of stress at point A on the
cross section at section a-a of the cantilever beam. Show the
results in a differential element at the point.
113-.
.13-4. Determine the magnitude of the load P that will
cause a maximum normal stress of Umax = 30 ksi in the link
along section a-a.
400kN
,a
.
•
l'a-o.sm-I
.-2in.p
a
a
,---~..L
I o.s in.
1-2in.-IT
I
p
lOOmm
_L
1---<
IOOmm
Section a-a
Prob F 3-2
Prob. F13-4
608
CHAPT ER
13
COMBIN ED LOADINGS
Fl3-5. The beam has a rectangular cross section and is
subjected to the loading shown. Determine the state of
stress at point B. Show the results in a differential element
at the point.
F13-7. Determine the state of stress at point A on the
cross section of the pipe at section a-a. Show the results in a
differential element at the point.
~
300mm
y
Section a - a
Prob. F13-5
Prob.Fl3-7
F13-6. Determine the state of stress at point A on the
cross section of the pipe assembly at section a-a. Show the
results in a differential element at the point.
F13-8. Determine the state of stress at point A on the
cross section of the shaft at section a-a. Show the results in
a differential element at the point.
300mm
x
~mm
lOOON
300 N 900 N ~mm
)'
w20mm
A
Section a - a
Prob. F13-6
Section a - a
Prob.Fl3-8
lOOmm
13.2
STATE OF STRESS CAUSED BY COMBINED LOADINGS
609
PROBLEMS
13-18. Determine the shortest distance d to the edge
of the plate at which the force P can be applied so that it
produces no compressive stresses in the plate at section a-a.
The plate has a thickness of 10 mm and P acts along the
centerline of this thickness.
13-21. If the load has a weight of (i()O lb, determine
the maximum normal stress on the cross section of
the supporting member at section a-a. Also, plot the
normal-stress distribution over the cross section.
I
300 mm
la
a
200mm-
K
0 - -1+--
0
lit)
Section o - o
500 mm
l
•
d
p
Prob.13-18
Prob. 13-21
13-19. Determine the maximum distanced to the edge of
the plate at which the force P can be applied so that it produces
no compressive stresses on the plate at section a-a. The plate
has a thickness of 20 mm and P acts along the centerline of
this thickness.
*13-20. The plate has a thickness of 20 mm and the force
P = 3 kN acts along the centerline of this thickness such that
d = 150 mm. Plot the distribution of normal stress acting along
section a-a.
200mm
13-22. The steel bracket is used to connect the ends of
two cables. If the allowable normal stress for the steel is
u allow = 30 ksi, determine the largest tensile force P that can
be applied to the cables. Assume the bracket is a rod having
a diameter of 1.5 in.
13-23. The steel bracket is used to connect the ends of
two cables. If the applied force P = 1.50 kip, determine the
maximum normal stress in the bracket. Assume the bracket
is a rod having a diameter of 1.5 in.
o
p
d
l
o
Probs.13-19/20
Probs. 13-22123
610
CHAPTER
13
CO MBIN ED LOADINGS
*13-24. The column is built up by gluing the two boards
together. Determine the maximum normal stress on the cross
section when the eccentric force of P = 50 kN is applied.
13-25. The column is built up by gluing the two boards
together. If the wood has an allowable normal stress of
u allow = 6 MPa, determine the maximum allowable eccentric
force P that can be applied to the column.
*13-28. The joint is subjected to the force system shown.
Sketch the normal-stress distribution acting over section a-a
if the member has a rectangular cross section of width 0.5 in.
and thickness 1 in.
13-29. The joint is subjected to the force system shown.
Determine the state of stress at points A and B , and sketch
the results on differential elements located at these points.
The member has a rectangular cross-sectional area of width
0.5 in. and th ickness 1 in.
p
250n~
~150mm
:>'
75mm
150 m.:;;;<
~
BCJ
<mm ~
1 in.
A
a-I
2in.
~
Probs. 13-24125
SOOlb
Probs. 13-28129
13-26. The screw of the clamp exerts a compressive force
of 500 lb on the wood blocks. Determine the maximum
normal stress along section a-a. The cross section is
rectangular, 0. 75 in. by 0.50 in.
13-30. The rib-joint pliers are used to grip the smooth
pipe C. If the force of 100 N is applied to the handles,
determine the state of stress at points A and B on the cross
section of the jaw at section a-a. Indicate the results on an
element at each point.
13-27. The screw of the clamp exerts a compressive
force of 500 lb on the wood blocks. Sketch the stress
distribution along section a-a of the clamp. The cross
section is rectangular, 0.75 in. by 0.50 in.
lOON
1 - - - - - 250 mm - - -- •
-I
A )'lOmm
4 in.
l
20
5:5~YJ.s
mm
Section a - a
Probs. 13-26/27
Prob.13-30
13.2
13-31. The t-in.-diameter bolt hook is subjected to the
load of F = 150 lb. Determine the stress components at
point A on the shank. Show the result on a volume element
located at this point.
611
STATE OF S TRESS CAUSED BY COMBINED LOADINGS
13-35. The spreader bar is used to lift the 2000-lb tank.
Determine the state of stress at points A and B. and indicate
the results on a differential volume element.
*13-32. The i -in.-diameter bolt hook is subjected to the
load of F = 150 lb. Determine the stress components at
point B on the shank. Show the result on a volume element
located at this point.
A
B~
-1(-
1 in.
l in.
1.5 in.
A
Prob. 13-35
l-2 in.-l- 2 in. -1 8
/
F=1501b
Probs. 13-31132
13-33. The block is subjected to the eccentric load shown.
Detennine the normal stress developed at points A and B.
Neglect the weight of the block.
13-34. The block is subjected to the eccentric load shown.
Sketch the normal-stress distribution acting over the cross
section at section a-a. Neglect the weight of the block.
*13-36. The drill is jammed in the wall and is subjected to
the torque and force shown. Determine the state of stress
at point A on the cross section of the drill bit at section a-a.
13-37. The drill is jammed in the wall and is subjected
to the torque and force shown. Determine the state of
stress at point B on the cross section of the drill bit at
section a-a.
y
~400mm--•
la
ISOkN
Section a - a
Probs. 13-33/34
Probs. 13-36/37
612
CHAPTER
13
CO MBIN ED LOADINGS
13-38. The frame supports the distributed load shown.
Determine the state of stress acting at point D. Show the
results on a differential element at this point.
13-39. The frame supports the distributed load shown.
Determine the state of stress acting at point £. Show the
results on a differential element at this point.
13-42. The beveled gear is subjected to the loads shown.
Determine the stress components acting on the shaft at
point A , and show the results on a volume element located
at this point. The shaft has a diameter of 1 in. and is fixed to
the wall at C.
13-43. The beveled gear is subjected to the loads shown.
Determine the stress components acting on the shaft at
point B, and show the results on a vol ume element located
at this point. The shaft has a diameter of 1 in. and is fixed to
the wall at C.
z
y
c
c
x
Probs. 13-38/39
8 in.
/
*13-40. The rod has a diameter of 40 mm. If it is subjected
to the force system shown, determine the stress components
that act at point A , and show the results on a volume element
located at this point.
13-41. Solve Prob. 13-40 for point B.
Probs. 13-42/43
*13-44. Determine the normal-stress developed at points
A and B. Neglect the weight of the block.
13-45. Sketch the normal-stress distribution acting over
the cross section at section a-a. Neglect the weight of the
block.
y
100mm
6 kip
3 in.
12 kip
z
B
1500N
600N
Probs. 13-40/41
Probs. 13-44/45
13.2
1~. The vertebra of the spinal column can support a
maximum compressive stress of <Tmax• before undergoing a
compression fracture. Determine the smallest force P that
can be applied to a vertebra if we assume this load is applied
at an eccentric distance e from the centerline of the bone,
and the bone remains elastic. Model the vertebra as a hollow
cylinder with an inner radius r, and outer radius r,,.
STATE OF S TRESS CAUSED BY COMBINED LOADINGS
613
13-50. The C-frame is used in a riveting machine. U the
force at the ram on the clamp at D is P = 8 kN. sketch the
stress distribution acting over the section a-a.
13-5L Determine the maximum ram force P that can be
applied to the clamp at D if the allowable normal stress for
the material is u allo..• = 180 MPa.
p
a-~ ~-u--a
p
D
-200 mmIOmm
~}omm
l, 1,1-
Prob.~
60 mm .I '-I 0 mm
Probs. 13-50/51
13-47. The solid rod is subjected to the loading shown.
Determine the state of stress at point A , and show the results
on a differential volume element located at this point.
*13-48. The solid rod is subjected to the loading shown.
Determine the state of stress at point 8. and show the results
on a differential volume element at this point.
13-49. The solid rod is subjected to the loading shown.
Determine the state of stress at point C. and show the results
on a differential volume element at this point.
*13-52. The uniform sign has a weight of 1500 lb and is
supported by the pipe AB, which has an inner radius of
2.75 in. and an outer radius of 3.00 in. lf the face of the sign
is subjected to a uniform wind pressure of p = 150 lb/ft2 ,
determine the state of stress at points C and D. Show the
results on a differential volume element located at each of
these points. Neglect the thickness of the sign. and assume that
it is supported along the outside edge of the pipe.
13-53.
Solve Prob. 13-52 for points£ and F.
y
200mm
150 lb/ft2
IA
x
B
E F
~o
z
A
z----- x1----Y
10 kN
Probs. 13-47/48149
Probs. 13-52153
614
CHAPTER
13
CO MBIN ED LOADINGS
CHAPTER REVIEW
A pressure vessel is considered to have a
thin wall provided r / r > 10. If the vessel
contains gas having a gage pressure p , then
for a cylindrical vessel, the circumferential
or hoop stress is
pr
U] -
I
This stress is twice as great as the
longitudinal stress,
pr
= 21
Thin-walled spherical vessels have the
same stress within their walls in all
directions. It is
U2
pr
U1
= U?- = -21
Superposition of stress components can be
used to determine the normal and shear
stress at a point in a member subjected to
a combined loading. To do this, it is first
necessary to determine the resultant axial
and shear forces and the resultant
torsional and bending moments at the
section where the point is located. Then
the normal and shear stress resultant
components at the point are determined
by algebraically adding the normal and
shear stress components of each loading.
VQ
-r= -
lt
Tp
J
7= -
M"
I
u= - '
CONCEPTUAL PROBLEMS
6 15
CONCEPTUAL PROBLEMS
C13-1. Explain why failure of this garden hose occurred
near its end and why the tear occurred along its length. Use
numerical values to explain your result. Assume the water
pressure is 30 psi.
C13-3. Unlike the turnbuckle at B, which is connected
along the axis of the rod, the one at A has been welded to the
edges of the rod, and so it will be subjected to additional stress.
Use the same numerical values for the tensile load in each rod
and the rod's diameter, and compare the stress in each rod.
Prob. C13-1
C13-2. This open-ended silo contains granular material. It
is constructed from wood slats and held together with steel
bands. Explain, using numerical values, why the bands are
not spaced evenly along the height of the cylinder. Also,
how would you find this spacing if each band is to be
subjected to the same stress?
Prob. C13-2
Prob. C13-3
C13-4. A constant wind blowing against the side of this
chimney has caused creeping strains in the mortar joints, such
that the chimney has a noticeable deformation. Explain how
to obtain the stress distribution over a section at the base of
the chimney, and sketch this distribution over the section.
Prob. C13-4
616
CHAPTER
13
CO MBIN ED LOADINGS
REVIEW PROBLEMS
R13-1. The post has a circular cross section of radius c.
Determine the maximum radius e at which the load P can
be applied so that no part of the post experiences a tensile
stress. Neglect the weight of the post.
R13-3. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of stress
at point Fon the cross section of the frame at section b-b.
Indicate the results on an element.
50mm
II
p
25mm[ ]
£ -1
75mm
_J_
Section a - a
tS1;
75mm
b
1-i
_J~_:=~~A~ [ ] 7f mm
1-1
Prob.Rl3-1
25mm
Section b - b
R13-2. The 20-kg drum is suspended from the hook
mounted on the wooden frame. Determine the state of stress
at point E on the cross section of the frame at section a-a.
Indicate the results on an element.
Prob. R13-3
*Rl3-4. The gondola and passengers have a weight of
1500 lb and center of gravity at G. The suspender arm AE
has a square cross-sectional area of 1.5 in. by 1.5 in., and is
pin connected at its ends A and £. Determine the largest
tensile stress developed in regions AB and DC of the arm.
50mm
II .£1-'
1-1
25 mm
75mm
_I
Section a - a
0.5
mm
0.5
B- ;11 1 m
11i·~7 )"
. - - c
1.5 in.-
5.5 ft
75mm
i-1
b
·A
_:z_.:::::::ill.'.~
[]1~mm
1-1-
25 mm
Section b - b
Prob.Rl3-2
G
Prob. R13-4
REVIEW PROBLEMS
R13-5. If the cross section of the femur at section a-a can
be approximated as a circular tube as shown, determine the
maximum normal stress developed on the cross section at
section a-a due to the load of 75 lb.
617
R13-7. The wall hanger has a thickness of 0.25 in. and is
used to support the vertical reactions of the beam that is
loaded as shown. If the load is transferred uniformly to
each strap of the hanger, determine the state of stress at
points C and D on the strap at A. Assume the vertical
reaction F at this end acts in the center and on the edge of
the bracket as shown.
10 kip
a
-
1-Hf - - O
Section a-a
F
F
Prob. R13-5
Rl.3-6. A bar having a squa re cross section of 30 mm by
30 mm is 2 m long and is held upward. U it has a mass of
5 kg/m, determine the largest angle 9. measured from the
vertical, at which it can be supported before it is subjected
to a tensile stress along its axis near the grip.
Prob. R13-6
Prob. R13-7
*Rl3-8. The wall hanger has a thickness of 0.25 in. and is
used to support the vertical reactions of the beam that is
loaded as shown. If the load is transferred uniformly to
each strap of the hanger, determine the state of stress at
points C and D on the strap at B. Assume the vertical
reaction F at this end acts in the center and on the edge of
the bracket as shown.
10 kip
Prob. R13-8
CHAPTER
14
These turbine blades are subjected to a complex pattern of stress. For design it is
necessary to determine where and in what direction the maximum stress occurs.
STRESS AND STRAIN
TRANSFORMATION
CHAPTER OBJECTIVES
•
To develop the transformation of stress and strain components from
one orientation of the coordinate system to another orientation.
•
To determine the principal stress and strain and the maximum
in-plane shear stress at a point.
•
To show how to use Mohr's circle for transforming stress and strain.
•
To discuss strain rosettes and some important material property
relationships, such as the elastic modulus, shear modulus, and
Poisson's ratio.
14.1
PLANE-STRESS TRANSFORMATION
It was shown in Sec. 7.3 that the general state of stress at a point is
characterized by six normal and shear-stress components, shown in
Fig. 14- la. This state of stress, however, is not often encountered in
engineering practice. Instead, most loadings are coplanar, and so the stress
these loadings produce can be analyzed in a single plane. When this is the
case, the material is then said to be subjected to plane stress.
u,
1
General state of stress
(a)
Fig. 14-1
619
620
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
u,
Plane stress
(b)
Fig. 14-1 (cont.)
)'
Uy
'T:ry
x
U.T
(a)
II
x'
(b)
Fig.14-2
The general state of plane stress at a point, shown in Fig. 14-lb, is
therefore represented by a combination of two normal-stress components,
a., ay, and one shear-stress component, 7:ry, which act on only four faces of
the element. For convenience, in this book we will view this state of stress
in the x- y plane, as shown in Fig. 14-2a. Realize, however, that if this state
of stress is produced on an element having a different orientation 6,
as in Fig.14-2b, then it will be subjected to three different stress components,
a,., ay•, Tr•y•, measured relative to the x' , y' axes. In other words, the state of
plane stress at the point is uniquely represented by two normal-stress
components and one shear-stress component acting on an element. To be
equivalent, these three components will be different for each specific
orientation (J of the element at the point.
If these three stress components act on the element in Fig. 14-2a, we
will now show what their values will have to be when they act on the
element in Fig.14-2b. This is similar to knowing the two force components
F, and Fy directed along the x, y axes, and then finding the force
components Ft' and Fy' directed along the x', y' axes, so they produce
the same resultant force. The transformation of force must only account
for the force component's magnitude and direction. The transformation
of stress components, however, is more difficult since it must account for
the magnitude and direction of each stress and the orientation of the
area upon which it acts.
14.1
PROCEDURE FOR ANAL YS/S
y
Uy
I
If the state of stress at a point is known for a given orientation of an
e leme nt, Fig. 14-3a, then the state of stress on an e lement having
some other orientation 8. Fig. 14-3b, can be determined as follows.
Try
Ur
-
• The normal and shear stress components ax· , 'x'y' acting on
the +x' face of the element, Fig. 14-3b, can be determined
from a n arbitra ry section of the e lement in Fig. 14-3a as
shown in Fig. 14-3c. If the sectioned area is ~ . then the
adjacent areas of the segment will be ~A sin 8 and ~ cos 8.
x face
(a)
II
• Draw the free-body diagram of the segment, which requires
showing the forces that act on the segment, Fig. 14-3d. This is
done by multiplying the stress components on each face by the
area upo n which they act.
y'
\
• When "i.F'.r· = 0 is applied to the free-body diagram, the area
~A will cancel out of each te rm and a direct solution for <T.r·
will be possible. Likewise, If'y· = 0 will yield ' x'y' ·
• If <ry· , acting o n the +y' face of the element in Fig. 14-3b, is to
be determined, then it is necessary to consider an arbitrary
segment of the element as shown in Fig. 14-3e. Applying
lf'y· = 0 to its free-body diagram will give a y'·
y'
(b)
y
y'\
\
x'
0
Try AA
x
y'
/
x'
u,·AA
u , AA cosO
\
u 'I'
y' face
/
x'
t-+-- u,
,.,,.AA cos 0
AA sin 0
'TxyAA sin 0
u 1 AA sin 9
(c)
(d)
Fig. 14-3
<Ty
(e)
621
PLAN E-STRESS TRANSFORMATION
X
622
I
CHAPTER
EXAMPLE
14
STRESS AND STRAIN TRANSFORMATI ON
14.1
The state of plane stress at a point on the surface of the airplane fuselage is
represented on the element oriented as shown in Fig. l4-4a. Represent the
state of stress at the point on an element that is oriented 30° clockwise from
this position.
1000
b
50MPa
0
a
t-"'MP•
>J'
b
'=::::'.::!:::::::;~.5 MPa
a
(a)
SOLUTION
~A
.O.A sin 30"~
.O.A cos 30"
(b}
If we apply the equations of force equilibrium in the x'
and y'directions, not the x and y directions, we will be able to obtain
direct solutions for u,. and Tr'y' ·
Equilibrium.
x'
k-x
y'~
The rotated element is shown in Fig.14-4d.To obtain the stress components
on this element we will first section the element in Fig. l4-4a
by the line a-a. The bottom segment is removed, and assuming the
sectioned (inclined) plane has an area ~A, the horizontal and vertical
planes have the areas shown in Fig. 14-4b. The free-body diagram of this
segment is shown in Fig. 14-4c. Notice that the sectioned x' face is defined
by the outward normal x' axis, and they' axis is along the face.
25 .O.A sin 30"
+ /'IF.x ·
=
O·'
O:t'~A
+
- (50 dA cos 30°) cos 30°
(25 dA .cos 30°) sin 30° + (80 dA sin 30°) sin 30°
+ (25 dA sin 30°) cos 30° = O
u r' =
-4.15 MPa
Ans.
r">"dA - (50 ~A cos 30°) sin 30°
- (25 ~A cos 30°) cos 30° - (80 ~A sin 30°) cos 30°
+ (25 dA sin 30°) sin 30° = O
300
50 .O.A cos 30°
(c)
Fig.14-4
'Tx'y' =
68.8 MPa
Ans.
Since <Tr' is negative, it acts in the opposite direction of that shown in
Fig. 14-4c. The results are shown on the top of the element in Fig. 14-4d,
since this surface is the one considered in Fig. 14-4c.
14.1
4.15 MPa
We must now repeat the procedure to obtain the stress on the
perpendicular plane b-b. Sectioning the element in Fig. 14-4a along b-b
results in a segment having sides with areas shown in Fig.14-4e. Orienting
the +x' axis outward, perpendicular to the sectioned face, the associated
free-body diagram is shown in Fig. 14-4f Thus,
+ 'i.Ifr·
= O;
a
ux·dA - (25 dA cos 30") sin 30"
25.SMPa
+ (80 dA cos 30°) cos 30" - {25 M sin 30°) cos 30"
- (50 M sin 30°) sin 30°
O'x ·
+.i"IF,,·
= O;
"x'y' dA
=
- 25.8 MPa
=0
b
{d)
Ans.
+ (25 dA cos 30°) cos 30°
+ (80 dA cos 30°) sin 30° - (25 dA sin 30°) sin 30°
+ (50 d A sin 30°) cos 30° = O
Tx'y'
=
- 68.8 MPa
An.1~
4A
Since both <Tx• and "~r· are negative quantities, they act opposite to their
direction shown in Fig. 14-4[ The stress components are shown acting on
the right side of the element in Fig. 14-4d.
From this analysis we may therefore conclude that the state of stress at
the point can be represented by a stress component acting on an element
removed from the fuselage and oriented as shown in Fig. 14-4a, or by
choosing one removed and oriented as shown in Fig. 14-4d. In other
words. these states of stress are equivalent.
25 4 A cos 30"
(f)
Fig. 14-4 (cont.)
6 23
PLANE-STRESS TRANSFORMATION
cos 300
(c)
624
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
14. 2
GENERAL EQUATIONS OF
PLANE-STRESS TRANSFORMATION
The method of transforming the normal and shear stress components
from the x, y to the x ', y' coordinate axes, as discussed in the previous
section, can be developed in a general manner and expressed as a set of
stress-transformation equations.
Uy
+ -rxy
Sign Convention. To apply these equations we must first establish a
+u x
--x
(a)
y
x'
u,· /
+o
sign convention for the stress components. As shown in Fig. 14-5, the +x
and +x ' axes are used to define the outward normal on the right-hand
face of the element, so that <T.r and <T.r' are positive when they act in the
positive x and x ' directions, and ?:ry and ?:r'y' are positive when they act in
the positive y and y' directions.
The orientation of the face upon which the normal and shear stress
components are to be determined will be defined by the angle 8, which
is measured from the +x axis to the +x' axis, Fig.14-5b. Notice that the
unprimed and primed sets of axes in this figure both form right-handed
coordinate systems; that is, the positive z (or z') axis always points out
of the page. The angle 8 will be positive when it follows the curl of the
right-hand fingers, i.e., counterclockwise as shown in Fig. 14-5b.
Normal and Shear Stress Components. Using this established
sign convention, the element in Fig. 14-6a is sectioned along the inclined
plane and the segment shown in Fig. 14-6b is isolated. Assuming the
sectioned area is ~A , then the horizontal and vertical faces of the
segment have an area of ~A sin 8 and ~A cos 8, respectively.
(b)
Positive sign convention
Fig.14-5
y
y'
\
1-+--•--x
fY.T
x'
llA cos 0
llA sin 0
(a)
(b)
Fig.14-6
14.2
The resulting free-body diagram of the segment is shown in Fig. 14-6c.
If we appl y the equatio ns o f equilibrium along the x ' and y' axes, we can
obtain a direct solution for <Tx· and rx'y'· We have
+ J'Il'.,.
=
O;
<Tx·
TxyilA
+ (rxyM
1'x·y•
T .r'y'
+ <Ty
2
=-
x'
= (<Ty -
+
<T.r - <Ty
2
=0
T.,. .lA COS 0
sin 8) sin 8 - (uyM sin 8 ) cos 8
0" 6
1
=0
(c)
2
u,.) sin 8 cos 8 + r.ry (cos2 8 - sin 8)
To simplify these two equations, use the trigonometric identities sin 28
2 sin 8 cos 8, s in2 8 = (1 - cos 28)/ 2, and cos2 8 = (1 + cos 28)/ 2.
Therefore,
<T.r· -
\
= <Tx cos2 8 + <Ty sin2 8 + T.ry(2 sin 8 cos 8)
- (':ryilA cos 8 ) cos 8 + (ux M cos 8) sin 8
<T.r
y'
- (r.rydA sin 8) cos 8 - (uyM sin 8) sin 8
<Tx·M
- (rxyilA cos 8) sin 8 - (ux M cos 8) cos 8
+'IF,.· = O;
<T, - <Ty
cos 28
·
2
+ 1'xy sin 28
sin 28 + r,.y cos 28
=
y'
(14-1)
(14-2)
Stress Components Acting alo ng x'. y' Axes
(d)
If the normal stress acting in they' direction is needed, it can be obtained
by simply substituting 8 + 90° for 8 into Eq. 14-1, Fig. 14-6d. This yields
<Ty• =
<Tx - <Ty
2
625
GENERAL EQUATIONS O F PLAN E-STRESS TRANSFORMATION
2
COS 28
.
- Txy Sill 28
(14-3)
PROCEDURE FOR ANALYSIS
To apply the stress transformation Eqs. 14-1 and 14-2, it is simply
necessary to substitute in the known data for ux, <Ty, TxY' and () in
accordance with the established sign convention, Fig. 14-5.
Remember that the x' axis is always directed positive outward from
the plane upon which the normal stress is to be determined. The
angle 8 is positive counterclockwise, from the x to the x' axis. If <T.r·
and ':r'y' are calculated as positive quantities, then these stresses act
in the positive direction of the x' and y' axes.
For convenience, these equations can easily be programmed on a
pocket calculator.
Fig. 1~
A sin 0
626
14
CHAPTER
EXAMPLE
STRESS AND STRAIN TRANSFORMATI ON
14.2
-
~
The state of plane stress at a point is represented on the element shown in
Fig. 14-7a. Determine the state of stress at this point on another element
oriented 30° clockwise from the position shown.
50 MPa
'
t-80MP•
-
25MPa
To obtain the stress components on plane CD, Fig. 14-7b, the
positive x' axis must be directed outward, perpendicular to CD, and the
associated y ' axis is directed along CD. The angle measured from the x to
the x ' axis is 6 = -30° (clockwise). Applying Eqs. 14-1 and 14-2 yields
<Tr + <Tv
<Tr - <Tv
<T • =
·
· +
· 2 · cos26 + Trn
x
2
.., sin26
y'
!
() = - 300
r
This problem was solved in Example 14.1 using basic principles. Here
we will apply Eqs. 14- 1 and 14- 2. From the established sign convention,
Fig. 14- 5, it is seen that
<Tr = -80 MPa
<Ty = 50 MPa
'Try = - 25 MPa
Plane CO.
(a)
30•
SOLUTION
D
x
x'
=
-80 + 50
-80 - 50
+
cos 2(-30°) + (-25) sin 2(-30°)
2
2
-25.8MPa
Ans.
(b}
<T.t -
'T<'y'
x'
=
y'
~B
=
() =600
x
c
<Ty
.
'Try cos 26
2
-80 - 50
sin 2(-30°) + (-25) cos 2(-30°)
2
-68.8MPa
= -
sin 26 +
Ans.
The negative signs indicate that <T.t ' and -i:t'y' act in the negative x' and y '
directions, respectively. The results are shown acting on the element in
Fig.14- 7d.
Establishing the x ' axis outward from plane BC, Fig. 14- 7c,
then between the x and x ' axes, 6 = 60° (counterclockwise). Applying
Eqs.14-1and14- 2,* we get
Plane SC.
(c)
4.15 MPa
~
<Tt'
=
-80 2+ 50 + -80 2- 50 cos 2(600) + (-25) sin 2(600)
=
-4.15 MPa
Ans.
-80 - 50
sin 2(60°) + (-25) cos 2(60°)
2
68.8MPa
Ans.
-i:t'y' = -
=
25.8MPa
68.8 MPa
(d)
Fig. 14-7
Here T<' y' has been calculated twice in order to provide a check. The
negative sign for u" indicates that this stress acts in the negative x '
direction, Fig. 14-7c. The results are shown on the element in Fig. 14-7d.
*Alternatively, we could apply Eq. 14-3 with
(I =
- 300 rather than Eq. 14-1.
14.3
627
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
14.3 PRINCIPAL STRESSES AND
MAXIMUM IN-PLANE SHEAR
STRESS
Since ux, u,,, T:ry are all constant, then from Eqs. 14-1 and 14-2 it can be
seen that the magnitudes of u x· and "x'y' only depend on the angle of
inclination 8 of the planes on which these stresses act. In engineering
practice it is often important to determine the orientation that causes the
normal stress to be a maximum, and the orientation that causes the shear
stress to be a maximum. We will now consider each of these cases.
In-Plane P incip I Stresses. To determine the maximum and
minimum normal stress, we must differentiate Eq. 14-1 with respect to 0
and set the result equal to zero. This gives
du..-
-- =
d8
U.r -
Uy
2
(2 sin 20) + 2-rxy cos 20
Solving we obtain the orientation 0
minimum normal stress.
tan 28"
=
0
= O" of the planes of maximum and
= (Ux
-
Uy
)/2
(14-4)
T
J(Ux;
Orientation of Principal Planes
The solution has two roots, Op, and OPl. Specifically, the values of 28p, and
20P2 are 180° apart, so Op, and 8"2 will be 90° apart.
I
Uyr+T.ry2
T .ry
~,..,.~~~t:;:.t"'-"--'--'-----'--'-l~ u
fig. 14--8
I
I
I
The cracks in this concrele beam
were caused by 1ension slrcss, even
though the beam was subjeclcd to
both an inte rnal momenl and shear.
The stress 1ransforma1ion equal ions
can be used 10 predicl the direclion
of the cracks, and the principal
normal stresses that caused !hem.
628
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
To obtain the maximum and minimum normal stress, we must
substitute these angles into Eq. 14-1. Here the necessary sine and cosine
of 28p, and 21JP2 can be found from the shaded triangles shown in
Fig. 14-8, which are constructed based on Eq. 14-4, assuming that 7:ry and
(a:, are both positive or both negative quantities.
After substituting and simplifying, we obtain two roots, a 1 and a 2 .
They are
ay)
_ a, + ay + \j/(a, - ay)2 +
a1.2 -
2
2
?:ry
2
(14-5)
Principal Stresses
These two values, with a 1 > a 2, are called the in-plane principal stresses,
and the corresponding planes on which they act are called the principal
planes of stress, Fig. 14-9. Fmally, if the trigonometric relations for 8p, or
8P 2 are substituted into Eq. 14-2, it will be seen that 7:r'y' = O; in other
words, no shear stress acts on the principal planes, Fig. 14-9.
'Txy
- "·'
-
II
In-plane principal stresses
Fig.14-9
14.3
629
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
Maximum In Plane Shear Stress. The orientation of the element
that is subjected to maximum shear stress can be determined by taking
the derivative of Eq. 14-2 with respect to 8, and setting the result equal to
zero. This gives
- (ax - a,.)/2
{14-6)
tan 28, =
Txy
T
-,....,
I
Orientation of Maximum In-Plane Shear Stress
The two roots of this equation, 8, 1 and 8,1 , can be determined from the
shaded triangles shown in Fig.14-lOa.Since tan 28...,Eq.14-6,is the negative
reciprocal of tan 28P, Eq. 14-4, then each root 20s is 90° from 28P,
and the roots 8, and (JP are 45° apart. Therefore, an element subjected to
maximum shear stress must be oriented 45° from the position of an
element that is subjected to the principal stress.
The maximum shear stress can be found by taking the trigonometric
values of sin 28, and cos 28s from Fig. 14-10 and substituting them into
Eq. 14-2. The result is
(a)
x'
-~
I
e,,
~~~~~~~~~~~~~
T
••.,,b.. =
111:u1
~(
<7.r -
2
Uy
2
)
+ 1:ry. 2
{14-7)
Maximum In-Plane Shear Stress
Here r:::;..,. is referred to as the maximum in-plane shear stress, because
it acts on the element in the x-y plane.
Finally, when the values for sin 28, and cos 28, are substituted into
Eq. 14-1, we see that there is also an average normal stress on the planes
of maximum in-plane shear stress. It is
Uavg
=
2
{14-8)
Average Normal Stress
For numerical applications, it is suggested that Eqs. 14-1through14-8 be
programmed for use on a pocket calculator.
IMPORTANT POINTS
• The principal stresses represent the maximum and minimum
normal stress at the point.
• When the state of stress is represented by the principal stresses,
no shear stress will act on the element.
• The state of stress at the point can also be represented in terms
of the maximum in-plane shear stress. In this case an average
normal stress will a lso act on the element.
• The e lement representing the maximum in-plane shear stress
with the associated average normal stresses is oriented 45°
from the e lement representing the principal stresses.
Maximum in-plane shear stresses
(b)
Fig. 14-10
630
CHAPTER
EXAMPLE
14
STRESS AND STRAIN TRANSFORMATI ON
14.3
-
~
The state of stress at a point jus~ before failure of this shaft is shown in
Fig.14-lla. Represent this state of stress in terms of its principal stresses.
SOLUTION
From the established sign convention,
ux = -20MPa
60 MPa
Try =
Applying Eq. 14-4,
Orientation of Element.
tan 28
90MPa
<ry =
-:i:rv
=
,
(ur - <ry)/2
P
60
(-20 - 90)/2
= ------
Solving, and referring to this first angle as Op,, we have
90MPa
2op, = -47.49°
--+----+ 60 MPa
op, = -23.7°
Since the difference between 28p, and 20p, is 180°, the second angle is
20p, = 180° + 20p, = 132.51°
Op, = 66.3°
i.+-- 20MPa
In both cases, 0 must be measured positive counterclockwise from the
x axis to the outward normal (x' axis) on the face of the element, and so
the element showing the principal stresses will be oriented as shown in
Fig. 14-llb.
(a)
Principal Stress.
We have
x'
<Tt,2 =
y'
~
2
<Ty
+
f
\J
=
-20 2+ 90 +
=
35.0 + 81.4
(u• 2
<Ty )2
+
~(-20 2- 90)
2
-i:ry
2
+ (60)2
u 1 = 116 MPa
x'
(b)
u, +
Ans.
-46.4 MPa
Ans.
The principal plane on which each normal stress acts can be determined
by applying Eq. 14- 1 with, say, 0 = Op, = -23.7°. We have
u2 =
u 1 =116 MPa
<Tr
fJP, = 23.7°
u 2 = 46.4 MPa
(c)
Fig.14-11
+ <Ty
<Tr -
<Tv
, cos 20 + Try sin 20
2
2
-20 + 90
-20 - 90
cos 2(-23.7°) + 60 sin 2(-23.7°)
2
+
2
- -46.4MPa
<Tr• =
+ .
Hence, u 2 = -46.4 MPa acts on the plane defined by Op, = -23.7°,
whereas u 1 = 116 MPa acts on the plane defined by Op, = 66.3°,
Fig. 14-llc. Recall that no shear stress acts on this element.
14.3
EXAMPLE
6 31
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
14.4
The state of plane stress at a point on a body is represented on the
element shown in Fig. 14-12a. Represent this state of stress in terms of its
maximum in-plane shear stress and associated average normal stress.
90MPa
----...-~
SOLUTION
t-WMP•
Orientation of Element. Since Ux = -20 MPa, Uy = 90 MPa, and
Txy = 60 MPa, applying Eq. 14-6, the two angles are
=
tan 26
'
-(a:, - uy)/2
-(-20 - 90)/2
= -
- - -- -
60
'Txy
26.2
= 42.5°
2fls,
= 180° +
60 MPa
6s, = 21.3°
(a)
851 = 111.3°
2Bs,
Note how these angles are formed between the x and x' axes, Fig. 14-12b.
They happen to be 45° away from the principal planes of stress, which
were determined in Example 14.3.
Maximum In-Plane Shear Stress.
7 ~-
= )( Ux
=
~Uy
Applying Eq.14-7,
2
)
+
'Tx/
= ) ( -20 2
2
90) + (60)2
+81.4 MPa
Ans.
The proper direction of T:;._ on the element can be determined by
substituting 6 = 6,, = 21.3° into Eq. 14-2. We have
'Tx•y•
= -( Ux ;
Uy) Sin
28 +
(b)
T_..y COS 28
-20 - 90) sin 2(21.3°) + 60 cos 2(21.3°)
2
81.4 MPa
35MPa
= -(
=
This positive result indicates that T::'.',:i- = Tx'y' acts in the positive y'
direction on this face (6 = 21.3°), Fig. 14-12b. The shear stresses on the
other three faces are directed as shown in Fig.14-12c.
Besides the maximum shear stress, the
element is also subjected to an average normal stress determined from
Eq. 14-8; that is,
Ux + Uy
-20 + 90
O'avg =
=
= 35 MPa
Ans.
Average Normal Stress.
2
2
This is a tensile stress. The results are shown in Fig. 14-12c.
(c)
Fig.14-12
632
I
CHAPTER
EXAMPLE
14
14.s
STRESS AND STRAIN TRANSFORMATION
I
When the torsional loading T is applied to the bar in Fig. 14-13a, it
produces a state of pure shear st ress in the material. Determine (a) the
maximum in-plane shear stress and the associated average normal stress,
and (b) the principal stress.
SOLUTION
From the established sign convention,
f01
ax = 0
= 0
O'.y
Maximum In-Plane Shear Stress.
(a)
'T!"""
m•plan~
=
aavg =
~(
Applying Eqs.14-7and14-8, we have
?
ar-ay -
. 2
)
a,+ av
' =
0
+
+
'T 2
xy
, /
= v(0)2 + (-T)2 = + 'T
Ans.
0
= 0
Ans.
2
2
Thus, as expected, the maximum in-plane shear stress is represented by
the element in Fig.14-13a.
NOTE: Through experiment it has been found that materials that are
Torsion failure of mild steel.
x'
ductile actually fail due to shear stress. As a result, if the bar in Fig. 14-13a
is made of mild steel, the maximum in-plane shear stress will cause it to fail
as shown in the adjacent photo.
Applying Eqs. 14-4 and 14-5 yields
Principal Stress.
'Try
'<::-----'--x
tan 28P = (
<T.r -
a1 ,2
(b)
=
<T.r: a y
+
ay
(<T.r
-T
)
/2
- (O _ 0)/ , 8p, = 45°, 8p, = -45°
2
~ ay )2 + 'Try2 =
0 + Y(0)2 +
'T2
=
Ans.
+'T
If we now apply Eq.14-1with8p, = 45°, then
Fig.14-13
= 0 + 0 + (-T) sin 90° =
-T
Thus, a 2 = -T acts at 8p, = 45° as shown in Fig. 14-13b, and a 1 =
on the other face, 8p, = -45°.
T
acts
NOTE: Materials that are brittle fail due to normal stress. Therefore, if the
bar in Fig. 14-13a is made of cast iron it will fail in tension at a 45°
inclination as seen in the adjacent photo.
Torsion failure of cast iron.
14.3
EXAMPLE
633
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
14.6
-
-
When the axial loading Pis applied to the bar in Fig. 14-14a, it produces a
tensile stress in the material. Determine (a) the principal stress and (b) the
maximum in-plane shear stress and associated average normal stress.
p
SOLUTION
From the established sign convention,
a, = a
ay = 0
By observation, the element oriented as shown in
Fig. 14-14a illustrates a condition of principal stress since no sh.ear stress
acts on this element. This can also be shown by direct substitution of the
above values into Eqs. 14-4 and 14-5. Thus,
Principal Stress.
a1 = a
(a)
Ans.
NOTE: Brittle materials will fail due to normal stress, and therefore, if the
bar in Fig. 14-14a is made of cast iron, it will fail as shown in the adjacent
photo.
Applying Eqs. 14-6, 14-7, and 14-8,
Maximum In-Plane Shear Stress.
we have
Axial failure of cast iron.
tan 26S =
T
!Dax
=
tn· plane
-(a, - ay)/2
-r:\')'
f(a, - ay)2 +
2
\j
-(a - 0)/ 2
=
0
T.
2=
-'Y
'
6S1 = 45° ' 6Si = -45°
\Jf(a 2-
0)2 + (0)2 = + a
- 2
a,+ ay
a +O a
aavg =
2
2
= z
Ans.
Ans.
To determine the proper orientation of the element, apply Eq. 14- 2.
T.• ,•
.t)
= -
~-~.
Sill
2
26 +
a-0.
0
T. COS 26 = ·'Y
2 Sill 90
+
a
0 = - 2
(b}
Fig.14-14
This negative shear stress acts on the x ' face in the negative y' direction,
as shown in Fig. 14-14b.
NOTE: If the bar in Fig. 14-14a is made of a ductile material such as mild
steel then shear stress will cause it to fail. This can be noted in the adjacent
photo, where within the region of necking, shear stress has caused
"slipping" along the steel's crystalline boundaries, resulting in a plane of
failure that has formed a cone around the bar oriented at approximately
45° as calculated above.
Axial failure of mild steel.
634
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
PRELIMINARY PROBLEMS
P14-L In each case, the state of stress u_,, u,. ,,._.,.produces
normal and shear stress components along section AB of the
element that have values of u _.• = - 5 kPa and,,._..,.. = 8 kPa
when calculated using the stress transformation equations.
Establish the x' and y' axes for each segment and specify the
angle IJ, then show these results acting on each segment.
A
A
B
(c)
Prob. P14-1
A
'-------\B
I
B
P14-2. Given the state of stress shown on the element,
find u avg and Tmax
and show the results on a properly
.
in-plane
oriented element.
(a)
rMPa
!Uy
I
B
30"\
.. ~·
T xy
;.t
•
B
4MPa
B
1
f
(b)
Prob. P14-2
14.3
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
635
FUNDAMENTAL PROBLEMS
F14-1. Determine the normal stress and shear stress
acting on the inclined plane AB.
F14-4. Determine the equivalent state of stress on an
element at the same point that represents the maximum
in-plane shear stress at the point.
700kPa
~B
---A~~JQ°
I
500kPa
II---••
100 kPa
Prob. F14-1
400kPa
Prob.Fl4-4
F14-2. Determine the equivalent state of stress on an
element at the same point oriented 45° clockwise with
respect to the element shown.
Fl4-S. The beam is subjected to the load at its end.
Determine the maximum principal stress at point B.
__LPa
101
T
300 kPa
Prob. F14-2
2kN
Prob.Fl4-S
F14-3. Determine the equivalent state of stress on an
element at the same point that represents the principal stresses
at the point 1. Also, find the corresponding orientation of
the element with respect to the element shown.
8kN/m
i - - - - -3 m - - - - i- - - - 3 n1 - - -
Prob. F14-3
Prob.Fl4-6
636
CHAPT ER
14
STR ESS AND STRAIN TRANSFORMATION
PROBLEMS
14-1. Prove that the sum of the normal stresses
ux + u,. = ux· + u1.. is constant. See Figs. 14-2a and 14-2b.
14-2. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 14.1.
*14-4. Determine the normal stress and shear stress
acting on the inclined plane AB. Solve the problem using
the method of equilibrium described in Sec. 14.1.
14-5. Determine the normal stress and shear stress acting
on the inclined plane AB. Solve the problem using the stress
transformation equations. Show the results on the sectional
element.
65MPa
A
15 ksi
l
3Cf
B
L l
T
B
6Cf
20MPa
6 ksi
Prob.14-2
Probs. 14-4/5
14-3. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 14.1.
A
14-6. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 14.1.
400 psi
650 psi
--t
---!-
8 ksi
5 ksi
4Cf
A
B
Prob.14-3
B
3 ksi
Prob.14-6
14.3
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
14-7. Dete rmine the stress components acting on the
inclined plane A 8. Solve the problem using the method of
equilibrium described in Sec. 14.1.
637
14-11. Determine the equivalent state of stress on an
element at the same point oriented 60° clockwise with respect
to the element shown. Sketch the results on the element.
*14-8. Solve Prob. 14-7 using the stress-transformation
equations developed in Sec. 14.2.
A
100 MPa
60MPa
75MPa
150 MPa
'
B
Prob. 14-11
Probs. 14-7/8
14-9. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 14.1.
14-10. Solve Prob. 14-9 using the stress-transformation
equation developed in Sec. 14.2.
*14-U. Determine the equivalent state of stress on an
element at the same point oriented 60° counterclockwise
with respect to the element shown. Sketch the results on the
element.
A
80MPa
100 MPa
75 MPa
150 MPa
40MPa
B
Probs. 14-9/ 10
Prob. 14-12
638
CHAPTER
14
STRESS AND S TRA IN TRANSFORMATION
14-13. Determine the stress components acting on the
inclined plane AB. Solve the problem using the method of
equilibrium described in Sec. 14.1.
SOMPa
A
*14-16. Determine the equivalent state of stress on an
element at the point which represents (a) the principal
stresses and (b) the maximwn in-plane shear stress and the
associated average normal stress. Also, for each case,
determine the corresponding orientation of the element
with respect to the clement shown and sketch the results on
the element.
IOOMPa
SOM Pa
ISMPa
B
Prob. 14-16
Prob. 14-13
14-14. Determine (a) the principal stresses and (b) the
maxin1wn in-plane shear stress and average normal stress at
the point. Specify the orientation or the element in each case.
200 MPa
14-17. Detem1inc the equivalent state or stress on an element
at the san1e point which represents (a) the principal stress. and
(b) the maximum in-plane shear stress and the associated
average normal stress. Also, for each case, determine the
corresponding orientation of the clement with respect to the
element shown and sketch the results on each element.
,.....=:::::!t=Z.., I00 MPa
H--•
75 MPa
300MPa
'
I 25MPa
SOMPa
Prob. 14-14
14-15. The state of stress at a point is shown on the element.
Determine (a) the principal stresses and (b) the maximum
in-plane shear stress and average normal stress at the point.
Specify the orientation of the element in each case.
60MPa
+
Prob. 14-17
14-18. A point on a thin plate is subjected to the two
stress components. Determine the resultant state of stress
represented on the element oriented as shown on the
right.
30 MPa
1-1--•
45 MPa
--'--Txy
85 MPa
Prob. 14-15
Prob. 14-18
14.3
PRINCIPAL STRESSES AND M AXIMUM IN-PLANE SHEAR STRESS
14-19. Determine the equivalent state of stress on an
element at the same point which represents (a) the principal
stress, and (b) the maximum in-plane shear stress and the
associated average normal stress. Also. for each case,
determine the corresponding orientation of the element
with respect to the element shown and sketch the results on
the element.
639
14-2L The stress acting on two planes at a point is
indicated. Determine the shear stress on plane a-a and the
principal stresses at the point.
b
a
~---~
25 MPa
60 ksi
b
Prob.14-19
Prob.14-21
*14-20. The stress along two planes at a point is indicated.
Determine the normal stresses on plane ~ and the
principal stresses.
14-22. The state of stress at a point in a member is shown
on the element. Determine the stress components acting on
the plane AB.
b
A
SOMPa
45MPa
- .------a
t1 - - - - - -.....
b
Prob.14-20
B
Prob.14-22
640
CHAPTER
14
STRESS ANO STRAIN TRANSFORMATION
The following problems involve material covered in
Chapter 13.
14-23. The wood beam is subjected to a load of 12 kN. If
grains of wood in the beam at point A make an angle of 25°
with the horizontal as shown, determine the normal and shear
stress that act perpendicular to the grains due to the loading.
14-27. A rod has a circular cros.s section with a diameter of
2 in. It is subjected to a torque of 12 kip · in. and a bending
moment M. The greater principal stress at the point of
maximum flexural stress is 15 ksi. Determine the magnitude
of the bending moment.
12kN
1-2 m-i-1 m•- t + -- 4 m.-
-11
D~mm
1-1
Prob. 14-27
200mm
Prob. 14-23
*14-28. The bell crank is pinned at A and supported by a
short link BC. If it is subjected to the force of 80 N, determine
the principal stresses at (a) point D and (b) point £. l11e
crank is constructed from an aluminum plate having a
thickness of 20 mm.
*14-24. l11e internal loadings at a section of the beam
are shown. Determine the in-plane principal stresses at
point A. Also compute the maximum in-plane shear stress
at this point.
50mm
D
150
mm=-! s
-y--t~~~:4~0Jh~1Dl;~~o~m:m:::::::::=:::===-si
50mm
14-25. Solve Prob. 14-24 for point B.
I
A
14-26. Solve Prob. 14-24 for point C.
I
50mm
E
15 mm
~20mm
Prob. 14-28
200mm
14-29. The beam has a rectangular cros.s section and is
subjected to the loadings shown. Determine the principal
stresses at point A and point B , which are located just to the
left of the 20-kN load. Show the results on elements located
at these points.
20kN
80kN
'-x
lOOmm
+·B
111
o--2m~-A-Probs. 14-24125126
..............
Prob. 14-29
10 kN 8
A
0::r
100
:::J: I00
!rl.!
50mm 50mm
nun
mm
14.3
PRINCIPAL STRESSES AND MAXIMUM IN-PLANE SHEAR STRESS
14-30. A paper tube is formed by rolling a cardboard strip
in a spiral and then gluing the edges together as shown.
Determine the shear stress acting along the seam, which is
at 500 from the horizontal. when the tube is subjected to an
axial compressive force of 200 N. The paper is 2 mm thick
and the tube has an outer diameter of 100 mm.
641
14-33. Determine the principal stresses in the cantilevered
beam at points A and 8.
14-3L Solve Prob. 14-30 for the normal stress acting
perpendicular to the seam.
100mm
Prob. 14-33
Probs. 14-30/31
*14-32. The 2-in.-diameter drive shaft AB on the
helicopter is subjected to an axial tension of 10 000 lb and a
torque of 300 lb· ft. Determine the principal stresses and
the maximum in-plane shear stress that act at a point on the
surface of the shaft.
14-34. The internal loadings at a cross section through the
6-in.-diameter drive shaft of a turbine consist of an axial
force of 2500 lb. a bending moment of 800 lb· ft, and a
torsional moment of 1500 lb · fl. Determine the principal
stresses at point A. Also calculate the maximum in-plane
shear stress at this point.
14-35. The internal loadings at a cross section through the
6-in.-diameter drive shaft of a turbine consist of an axial
force of 2500 lb, a bending moment of 800 lb· ft, and a
torsional moment of 1500 lb · ft. Determine the principal
stresses at point B. Also calculate the maximum in-plane
shear stress at this point.
~1500 lb·ft
Prob. 14-32
ProbS- 14-34/35
642
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
*14-36. The shaft has a diameter d and is subjected to the
loadings shown. Determine the principal stresses and the
maximum in-plane shear stress at point A. The bearings
only support vertical reactions.
14-39. The wide-flange beam is subjected to the 50-kN
force. Determine the principal stresses in the beam at point A
located on the web at the bottom of the upper flange.
Although it is not very accurate, use the shear formula to
calculate the shear stress.
*14-40. Solve Prob. 14-39 for point B located on the web
at the top of the bottom flange.
p
!
~~
1- j~
L
2
~~
L
2
SOkN
~
!
A
B
-
Prob.14-36
1 m -Ji-----3 m -----1
I
A
10 mm
I
125012mm
mm
I-I t 12mm
14-37. The steel pipe has an inner diameter of2.75 in. and
an outer diameter of 3 in. If it is fixed at C and subjected to
200mm
the horizontal 60-lb force acting on the handle of the pipe
wrench at its end, determine the principal stresses in the pipe
at point A , which is located on the outer surface of the pipe.
14-38. Solve Prob. 14-37 for point B, which is located on
the outer surface of the pipe.
Probs. 14-39/40
14-41. The box beam is subjected to the 26-kN force that
is applied at the center of its width, 75 mm from each side.
Determine the principal stresses at point A and show the
results in an element located at this point. Use the shear
formula to calculate the shear stress.
14-42. Solve Prob. 14-41 for point B.
26kN
B
B
i - - - 2 n1 - - i - - - - - 3
n1 - - - - 1
130mm
17 1
c
)'
ro
.>
TBA T 1smm
mm-1..:'.l.QI 75 mm
r----1
x
150mm
Probs. 14-37/38
Probs. 14-41142
14.4
643
MOHR'S CIRCLE- PLANE STRESS
14.4 MOHR'S CIRCLE-PLANE STRESS
In this section, we will show how to apply the equations for plane-stress
transformation using a graphical procedure that is often convenient to use
and easy to remember. Furthermore, this approach will allow us to
"visualize" how the normal and shear stress components <T.t ' and -r,y vary
as the plane on which they act changes its direction, Fig. 14- 15a.
If we write Eqs. 14-1 and 14-2 in the form
<T.t' -
a
,
+a
y
)
(a,
a
y
)
cos
( 2
2
a,
-a
y
)
.
-r,y
+
sm
( 2
·
=
=
·
28
-
28 +
7:ry
?:ry
sin 28
"·" ~
'Tx•y•
()
(14-9)
cos 28
(14-10)
then the parameter 8 can be eliminated by squaring each equation and
adding them together. The result is
[ _(a, +2 ay)]2 + 2_ (<T.r -2 ay) + 2
2
a,.
Tx'y '
-
T xy
Finally, since a_,, a>., ?:ry are known constants, then the above equation
can be written in a more compact form as
(a_,• - Uavg) 2 +
Tx• y• 2 =
R2
(14-11)
where
(14-12)
If we establish coordinate axes, a positive to the right and -r positive
downward, and then plot Eq. 14- 11, it will be seen that this equation
represents a circle having a radius R and center on the a axis at
point C(aavg• 0), Fig. 14-15b. This circle is called Mohr's circle, because it
was developed by the German engineer Otto Mohr.
1 - - - - - "·' - - - - - 1
7
(b)
(a)
Fig.14-15
x'
644
CHAPTER
14
7 xy
STRESS AND STRAIN TRANSFORMATI ON
= -' x·y·
11 =
(
U.T
=
o•
x,x'
Ux•
(a)
Each point on Mohr's circle represents the two stress components <T.t'
and 'T.r'y' acting on the side of t he element defined by the outward x' axis,
when this axis is in a specific direction 8. For example, when x' is
coincident with the x axis as shown in Fig. 14-16a, then () = 0° and
a,. = a" 'T.t 'y' = T.ry· We will refer to this as the "reference point" A and
plot its coordinates A(an Try) , Fig. 14-16c.
Now consider rotating the x ' axis 90° counterclockwise, Fig. 14-16b.
Then <T.t ' = a>" Tt'y' = --r.ry· These values are the coordinates of point
G(ay,-T,y) on the circle, Fig. 14-16c. Hence, the radial line CG is 180°
counterclockwise from the radial "reference line" CA. In other words, a
rotation () of the x ' axis on th e element will correspond to a rotation 28
on the circle in the same direction.
As discussed in the following procedure, Mohr's circle can be used to
determine the principal stresses, the maximum in-plane shear stress, or
the stress on any arbitrary plane.
x'
(Ux; Uy)
Uy
y'
I)=
90°
I
20 =
180°1
Txy
cl
I
_l
1 - - - U avg - - + - - - - - -
7;ry
x
u.
-
0 = Cf
u,
(b)
(c)
Fig. 14-16
14.4
PROCEDURE FOR ANAL YS/S
u,,
The following steps are re quired to draw and use Mohr's circle.
Construction of the Circle.
• Establish a coordinate syste m such that the horizontal axis
represents the normal stress u , with positive to the right, and
the vertical axis re presents the shear stress T , with positive
down wards, Fig. 14-17a.•
• Using the positive sign convention for u.. , uy, 'T.ry, Fig. 14-17a,
plot the cente r of the circle C, which is located on the u axis at
a distance <Tavg = (O:r + u,,)/2 from the origin, Fig. 14-17a.
• Plot the " refere nce point" A having coordinates A(o:., 'Txy).
This point represents the normal and shear stress components
on the e le ment's right-hand vertical face, and since the x ' axis
coincides with the x axis, this represents(} = 0°, Fig. 14- 17a.
• Connect point A with the center C of the circle and determine
CA by trigonometry. This represents the radius R of the circle,
Fig. 14-17a.
• Once R has been dete rmined, sketch the circle.
Principal Stress.
• The principal stresses u 1 and u 2 (u1 =::: u 2) are the coordinates
of points B a nd D , whe re the circle intersects the u axis, i.e.,
whe re T = 0, Fig. 14-17a.
• These stresses act on planes defined by angles fJp, and 8Pi'
Fig. 14-l?b. One of these angles is represented on the circle as
28p,· It is measured from the radial reference line CA to line CB.
• Using trigonome try, de te rmine op, from the circle. Remember
that the directio n of rota tion 28P on the circle (here it happe ns
to be counte rclockwise) re presents the same direction of
rotatio n BP from the refere nce axis ( +x) to the principal plane
( +x ' ), Fig. 14-17b.*
Maximum In-Plane Shear Stress.
• The ave rage normal stress and maximum in-plane shear stress
compo ne nts are determined from the circle as the coordinates
of e ither point E or F, Fig. 14-17a.
• In this case the angles 8,, and 8,, give the orientation of the
planes tha t contain these components, Fig. 14- 17c. The angle
28.,, is shown in Fig. 14-17a and can be determined using
trigonometry. H ere the rotation happens to be clockwise, from
CA to CE, and so 8,, must be clockwise on the element,
Fig. 14-17c.•
645
MOHR' S CIRCLE-PLANE STRESS
T 1C)'
u
t---u.,., ----i F
u,
u,·
0 = (f
E
T
(a)
U2
x'
/
OP,
x
(b)
in-plane
~--,-- X
o,
'
(c)
Fig. 14-17
x'
646
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
Stresses on Arbitrary Plane.
• The normal and shear stress components u,. and Tr· v· acting on
a specified plane or x' axis, defined by the angle 8, Fig. 14- 17d,
can be obtained by finding the coordinates of point P on the
circle using trigonometiry, Fig.14-17a.
• To locate P, the known angle 8 (in this case counterclockwise),
Fig. 14-17d, must be measured on the circle in the same
direction 28 (counterclockwise) from the radial reference line
CA to the radial line CP, Fig. 14- 17a.*
*If the,,. axis were constructed posirive upwards, then the angle 2i1 on the circle
would be measured in the opposire direcrion to the orientation() of the x' axis.
F
1 - - -Uavg
Txy
ur
1 - - - - -u.,- - - - - <
l - - - - - -U.'C· - - - - - - 1
(a)
y'
\
u.-r· ~·"'
<T.r
()
(d)
Fig. 14-17 (cont.)
x
14.4
EXAMPLE
64 7
MOHR'S CIRCLE- PLANE STRESS
14.7
-
-
Due to the applied loading, the element at point A on the solid shaft in
Fig. 14-18a is subjected to the state of stress shown. Determine the principal
stresses acting at this point.
SOLUTION
Construction of the Circle. From Fig.14-18a,
u, = -12 ksi
<Ty =
0
T,y =
T
-6 ksi
The center of the circle is located on the <Taxis at the point
-12 + 0
.
a:avg =
= -6 ks1
2
The reference point A(-12, -6) and the center C(-6, 0) are plotted in
Fig. 14-18b. From the shaded triangle, the circle is constructedl having a
radius of
R
6) 2
Y(12 -
=
+
(6) 2 =
6 ksi
(a)
8.49 ksi
Principal Stress. The principal stresses are indicated by the coordinates
of points B and D. We have, for u 1 > u 2,
u1 = 8.49 - 6 = 2.49 ksi
u2 =
-6 - 8.49
=
Ans.
-14.5 ksi
Ans.
The orientation of the element can be determined by calculating the
angle 26p, in Fig. 14-18b, which here is measured counterclockwise from
CA to CD. It defines the direction 6p, of u 2 and its associated principal
plane. We have
26
=
tan- 1
6p,
=
22.5°
p,
6
= 45.0°
12 - 6
The element is oriented such that the x ' axis or u 2 is directed
22.5° counterclockwise from the horizontal (x axis), as shown in
Fig.14-1&.
2.49 ksi
1 - - -12 - - - 1
D
1 -6 -
B
-'--+---'·--'---"-c---+--+--- u (ksi)
14.Sksi~x·
22.s•
x
-r (ksi)
(b)
(c)
Fig. 14-18
648
CHAPTER
EXAMPLE
14
STRESS AND STRAIN TRANSFORMATI ON
14.8
-
~
The state of plane stress at a point is shown on the element in Fig. 14-19a.
Determine the maximum in-plane shear stress at this point.
90MPa
SOLUTION
~-~-~ 60MPa
--t
f-WMP•
Construction of the Circle.
ax
=
-20MPa
Uavg =
F
-20
+ 90
2
=
35 MPa
1
60
s1(
I .~~::-+-----1
-
-r:ry• = 60 MPa
Point C and the reference point A(-20, 60) are plotted. Applying the
Pythagorean theorem to the shaded triangle to determine the circle's
u (MPa) radius CA, we have
c
f
90MPa
ay =
The a , -r axes are established in Fig. 14-19b. The center of the circle C is
located on the a axis, at the point
(a)
35
From the problem data,
A l20
R
=
V(60) 2
+ (55) 2
=
81.4 MPa
.__.._£-=----'Maximum In-Plane Shear Stress. The maximum in-plane shear stress
and the average normal stress are identified by point E (or F) on the
circle. The coordinates of point £(35, 81.4) give
-r (MPa)
(b)
Uavg =
Ans.
35 MPa
Ans.
-rm""
= 81.4 MPa
in•plan~
y'
The angle 8, 1, measured counterclockwise from CA to CE, can be found
from the circle, identified as 28,,. We have
\
81.4 MPa
35 MPa
x'
21.3°
28, ,
=
tan-1(20
+ 35)
60
=
42.5 0
~--L--- x
8,, =
21.3°
Ans.
This counterclockwise angle defines the direction of the x' axis,
Fig. 14-19c. Since point E has positive coordinates, then the average
normal stress and the maximum in-plane shear stress both act in the
positive x' and y ' directions as shown.
(c)
Fig. 14-19
14.4
EXAMPLE
14.9
The state of plane stress at a point is shown on the element in Fig. 14-20a.
Represent this state of stress on an element oriented 30° counterclockwise
from the position shown.
12 kSI.
SOLUTION
8 ksi
Construction of the Circle. From the problem data,
6 ksi
u:c
= -8 ksi
u1
= 12 ksi
-rx y =
-6 ksi
The u and -r axes are established in Fig. 14-20b. The center of the circle C
is on the u axis at the point
(a)
-8 + 12
= 2 ksi
2
The reference point for 9 = 0° has coordinates A(-8, -6).
Hence from the shaded triangle the radius CA is
<Tavg
R
=
= Y(10)2 + (6) 2 = 11.66
Stresses on 30° Element. Since the element is to be
rotated 30° counterclockwise, we must construct a radial
line CP, 2(30°) = 60° counterclockwise, measured from
CA (9 = 00), Fig.14-20b.Thecoordinatesofpoint P(ux·, T:cy )
must then be obtained. From the geometry of the circle,
</>
= tan-•
649
MOHR'S CIRCLE-PLANE STRESS
1~ = 30.%
0
<Tx·
Tx•y•
T
I/! = 6Cf' - 30.%0 = 29.04°
= 2 - 11.66 cos 29.04° = -8.20 ksi
= 11.66 sin 29.04° = 5.66 ksi
(ksi)
(b)
Ans.
Ans.
These two stress components act on face BD of the element shown in
Fig. 14-20c, since the x' axis for this face is oriented 30° counterclockwise
from the x axis.
The stress components acting on the adjacent face DE of the element,
which is 60° clockwise from the positive x axis, Fig. 14-20c, are represented
by the coordinates of point Q on the circle. This point lies on the radial
line CQ, which is 180° from CP, or 120° clockwise from CA. The
coordinates of point Qare
<Tx·
1:r'y'
= 2 + 11.66 cos 29.04° = 12.2 ksi
= -(11.66 sin 29.04) = -5.66 ksi
NOTE: Here Tx•y• acts in the -y' direction, Fig.14-20c.
(i()O
12.2ksiy
Ans.
(check)
Ans.
x'
(c)
Fig. 14-20
650
CHAPT ER
14
STRESS AND STRAIN TRANSFORMATI ON
FUNDAMENTAL PROBLEMS
F14-7. Use Mohr's circle to determine the normal stress
and shear stress acting on the inclined plane AB.
F14-11. Determine the principal stresses at point A on
the cross section of the beam at section a-a.
1 -3000101
,a
Prob. F14-7
F14-8. Use Mohr's circle to determine the principal
stresses at the point. Also, find the corresponding orientation
of the element with respect to the element shown.
50
nu~l
T
J_
A
1500101
- - - • 30kPa
II
11
MkP•
30kN
1-1
500101
Section a-a
Prob.Fl4-11
Prob. F14-8
F14-9. Draw Mohr's circle and determine the principal
stresses.
F14-12. Determine the maximum in-plane shear stress at
point A on the cross section of the beam at section a-a,
which is located just to the left of the 60-kN force. Point A is
just below the flange.
!
Afu~
60kN
Prob. F14-9
a,
F14-10. The hollow circular shaft is subjected to the
torque of 4 kN · m. Determine the principal stresses at a
point on the surface of the shaft.
~0.501
I
101
1-100 0101-1
A
100101- -
Prob. F14-10
100101
1
1800101
:;;;;;;;;~..... ,
400101
I
l
10 0101
Section a- a
Prob. F14-12
I
14.4
651
MOHR' S CIRCLE-PLANE STRESS
PROBLEMS
14-43. Solve Prob. 14-2 using Mohr·s circle.
*14-44. Solve Prob. 14-3 using Mohr·s circle.
*14-52. Determine the equivalent state of stress if an
element is oriented 60° clockwise Crom the element shown.
14-45. Solve Prob. 14-6 using Mohr·s circle.
,.------. 65 ksi
14-46. Solve Prob. 14-10 using Mohr's circle.
l ___ I
14-47. Solve Prob. 14-15 using Mohr·s circle.
*14-48. Solve Prob. 14-16 using Mohr·s circle.
.__
14-49. Mohr's circle for the state of stress is shown in
Fig. 14-17a. Show that finding the coordinates of point
P (ux·· Tx•y•) on the circle gives the same value as the stress
transformation Eqs. 14-1 and 14-2.
14-50. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the clement in each case.
_.
Prob. 14-52
14-53. Draw Mohr's circle that describes each of the
following states of st rcss.
80MPa
2 ksi
600 psi
,.....=:=!:::=~
j
60 MPa
sOO psi
1
f
T
(a)
(c)
Prob. 14-53
Prob. 14-50
14-51. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the clement in each case.
(b)
14-54. Draw Mohr's circle that describes each of the
following states of stress.
12 ksi
200 psi.
-t
1--• 3 ksi
•
(a)
Prob. 14-51
.
(b)
Prob. 14-54
100 psi
652
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
14-55. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
14-58. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
200MPa
20MPa
lOOMPa
lSOMPa
- lOOMPa
-
40MPa
Prob.14-55
Prob.14-58
14-59. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
10 ksi
*14-56. Determine (a) the principal stress and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
~---''---~ 8 ksi
Prob.14-59
*14-60. Draw Mohr's circle that describes each of the
following states of stress.
2 ksi
800 psi
Prob.14-56
- -.. 60MPa
--
800 psi
8 ksi
14-57. Determine (a) the principal stresses and (b) the
maximum in-plane shear stress and average normal stress.
Specify the orientation of the element in each case.
t
(a)
50MPa
-
l
1_ _,___ 30 MPa
(b)
(c)
Prob.14-60
14-61. The grains of wood in the board make an angle of
20° with the horizontal as shown. Determine the normal
and shear stresses that act perpendicular and parallel to the
grains if the board is subjected to an axial load of 250 N.
I
300mm ----~~
-__,,.
250N . -rL - - - - - --+....::::,--_-_- __
(200
Prob.14-57
101
Prob.14-61
25mm
60 2n510m N
1 4.4
14-62. The post is fixed supported at its base and a horizontal
force is applied at its end as shown, determine (a) the
maximum in·plane shear stress developed at A and (b) the
principal stresses at A.
z
MOHR' S CIRCLE- PLANE STRESS
653
14-65. The frame supports the triangular distributed load
shown. Determine the normal and shear stresses at point D
that act perpendicular and parallel, respectively, to the
grains. The grains at this point make an angle of 35° with the
horizontal as shown.
14-66. The frame supports the triangular distributed load
shown. Determine the normal and shear stresses at point E
that act perpendicular and parallel, respectively, to the
grains. The grains at this point make an angle of 45° with the
horizontal as shown.
900 N/m
Prob.14-62
3m
14-63. Determine the principal stresses, the maximum
in·plane shear stress, and average normal stress. Specify the
orientation of the element in each case.
E
T-r•l~3..1.0-mm
1.5 m
_] @A
20MPa
_J_
O _ SOmm
-I 1-'
100mm
80MPa
Probs. 14-65/66
30 MPa
Prob.14-63
14-67. The rotor shaft of the helicopter is subjected to the
tensile force and torque shown when the rotor blades
provide the lifting force to suspend the helicopter at midair.
If the shaft has a diameter of 6 in., determine the principal
stresses and maximum in·plane shear stress at a point
located on the surface of the shaft.
*14-64. The thin.walled pipe has an inner diameter of 0.5 in.
and a thickness of 0.025 in. If it is subjected to an internal
pressure of 500 psi and the axial tension and torsional loadings
shown, determine the principal stress at a point on the surface
of the pipe.
20 lb· ft
50 kip
20 lb· ft
Prob. 14-64
Prob. 14-67
654
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
*14-68. The pedal crank for a bicycle has the cross section
shown. If it is fixed to the gear at B and does not rotate while
subjected to a force of 75 lb, determine the principal stresses
on the cross section at point C.
.4
14-73. If the box wrench is subjected to the 50 lb force,
determine the principal stresses and maximum in-plane
shear stress at point A on the cross section of the wrench at
section a-a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
14-74. If the box wrench is subjected to the 50-lb force,
determine the principal stresses and maximum in-plane
shear stress at point B on the cross section of the wrench at
section a-a. Specify the orientation of these states of stress
and indicate the results on elements at the point.
in.c1
¥ffi
I0.4 in.
.
_l0.4 in .
0.2 10. 1-1
0.3 in.
Prob.14-68
12 in.
14-69. A spherical pressure vessel has an inner radius of
5 ft and a wall thickness of 0.5 in. Draw Mohr's circle for the
state of stress at a point on the vessel and explain the
significance of the result. The vessel is subjected to an
internal pressure of 80 psi.
14-70. The cylindrical pressure vessel has an inner radius of
1.25 m and a wall thickness of 15 mm. It is made from steel
plates that are welded along the 45° seam. Determine the
normal and shear stress components along this seam if the
vessel is subjected to an internal pressure of 8 MPa.
~.,,.
Prob.14-70
~
a
a
50 lb
0.5 in.
~
Section a - a
Probs. 14-73/74
14-75. The post is fixed supported at its base and the
loadings are applied at its end as shown. Determine (a) the
maximum in-plane shear stress developed at A and (b) the
principal stresses at A.
z
I
14-71. Determine the normal and shear stresses at point D
that act perpendicular and parallel, respectively, to the grains.
The grains at this point make an angle of 30° with the
horizontal as shown. Point D is located just to the left of the
10-kN force.
900lb
*14-72. Determine the principal stress at point D , which is
located just to the left of the 10-kN force.
9 in.
lOkN
lOOmm
B
l-1m- l -lm
D
100 mm I
I 2m
01-1-j 300 mm-"=C'=0"'-,,_-1
lOOmm
Probs. 14-71172
Prob.14-75
14.5
14.5
ABSOLUTE MAXIMUM SHEAR STRESS
6 55
ABSOLUTE MAXIMUM SHEAR
STRESS
Since the strength of a ductile material depends upon its ability to resist
shear stress, it becomes important to find the absolute maximum shear
stress in the material when it is subjected to a loading. To show how this
can be done, we will confine our attention only to the most common case
of plane stress,* as shown in Fig. 14-21a. Here both a 1 and a 2 are tensile.
If we view the element in two dimensions at a time, that is, in the y-z,x- z,
and x- y planes, Figs. 14-21b, 14-21c, and 14-21d, then we can use Mohr's
circle to determine the maximum in-plane shear stress for each case. For
example, Mohr's circle extends between 0 and a 2 for the case shown in
Fig. 14- 21b. From this circle, Fig. 14-21e, the maximum in-plane shear
stress is re:;... = a 2 /2. Mohr's circles for the other two cases are also
z
x
x- y plane stress
shown in Fig. 14-21e. Comparing all three circles, the absolute maximum
shear stress is
~
~
(a)
(14-13)
u 1 and u 2 have
the same sign
It occurs on an element that is rotated 45° about they axis from the
element shown in Fig. 14- 21a or Fig. 14- 21c. It is this out of plane shear
stress that will cause the material to fail, not rr::-~•••.
(r.r'y)max
(
(;.r'()max
Absolute maximum
shear stress
z
z
•
y
(e)
•
'----------- Y
(b)
X ---------~
'----------- X
(c)
Fig.14-21
*The case for three-dimensional stress is discussed in books related to advanced mechanics
of materials and the theory of elasticity.
(d)
)
. I
Maxmmm m-p ane
shear stress
656
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
z
)(
~
"=:><
(7yz-)max
--+-__.
(;r()m•x
(;r.·)m3X
x
y
\_ Maximum in-plane and
absolute maximum shear stress
x- y plane stress
7
(a)
(b)
Fig.14-22
In a similar manner, if one of the in-plane principal stresses has the
opposite sign of the other, Fig. 14-22a, then the three Mohr's circles that
describe the state of stress for the element when viewed from each plane
are shown in Fig. 14-22b. Clearly, in this case
7aM
max
=
2
(14-14)
u 1 and u 2 have
opposite signs
Here the absolute maximum shear stress is equal to the maximum
in-plane shear stress found from rotating the element in Fig. 14- 22a,
45° about the z axis.
IMPORTANT POINTS
• If the in-plane principal stresses both have the same sign, the
absolute maximum shear stress will occur out of the plane and
has a value of T ~'::, = Umax/2. This value is greater than the
in-plane shear stress.
• If the in-plane principal stresses are of opposite signs, then the
absolute maximum shear stress will equal the maximum
in-plane shear stress; that is, T '"'
= (umax - CTmin)/2.
mu
14.5
ABSOLUTE M AXIMUM SH EAR STRESS
EXAM PLE 14.10
The point on the surface of the pressure ves.sel in Fig. 14-23a is subjected to
the state of plane stress. Dete rmine the absolute maximum shear stress at
this point.
16- j
16MPa
1-- - -32- - -- 1
(a)
T(MPa)
(b)
Fig. 14-23
SOLUTION
The principal stresses are u 1 = 32 MPa, u 2 = 16 MPa. If these stresses
are plotted along the u axis, the three Mohr's circles can be constructed
that describe the state o f stress viewed in each of the three perpendicular
planes, Fig. 14-23b. The largest circle has a radius of 16 MPa and describes
the state of stress in the plane only containing u 1 = 32 MPa, shown
shaded in Fig. 14-23a. An orientation of an element 45° within this plane
yields the sta te of absolute maximum shear stress and the associated
average normal stress, namely,
-
T•..
O"avg
This same result for
Eq.14-13.
=
Ans.
16 MPa
= 16 MPa
-· can be obtained from direct application of
T ...
0"1
32
=
32 - 16
2
= - = - = 16 MPa
Ans.
2
2
32 + 0
O"avg =
= 16 MPa
2
By comparison, the maximum in-plane shear stress can be determined
from the Mohr's circle drawn between u 1 = 32 MPa and u 2 = 16 MPa,
Fig. 14-23b. This gives a value of
T ...
-·
T-.
in.p111ni:
O"avg
=
8 MPa
= 32 +
16 = 24 MP
2
a
6 57
658
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
EXAMPLE 14.11
-
~
Due to an applied loading, an element at a point on a machine shaft is
subjected to the state of plane stress shown in Fig. 14-24a. Determine the
principal stresses and the absolute maximum shear stress at the point.
'===;-' 40 psi
SOLUTION
(a)
Principal Stresses.
The in-plane principal stresses can be determined from Mohr's
circle. The center of the circle is on the a axis at
aavg = (-20 + 0) /2 = -10 psi. Plotting the reference point
A(-20, -40), the radius CA is established and the circle is
drawn as shown in Fig. 14-24b. The radius is
A
R
7
V (20 -
10)2 + (40) 2
=
41.2 psi
The principal stresses are at the points where the circle
intersects the a axis; i.e.,
a1 = -10 + 41.2 = 31.2 psi
a 2 = -10 - 41.2 = -51.2 psi
From the circle, the counterclockwise angle 28, measured from CA to the
-a axis, ts
(psi)
(b)
)!'
31.2ps~
=
x'
28
51.2psi/
38.0"
=
tan - 1 (
40
)
20 - 10
=
76.0°
Thus,
fJ = 38.0°
This counterclockwise rotation defines the direction of the x' axis and a 2,
Fig. 14-24c. We have
/
a 1 = 31.2 psi a 2 = -51.2 psi
(c)
211 = 76.0" + 90" = 166°
Since these stresses
have opposite signs, applying Eq.14-14 we have
Absolute Maximum Shear Stress.
A
'T ,., =
ma.
U2
\..
= - 51.2
J
Tabs=
41.2
max
-r
(psi)
(d)
Fig. 14-24
Ans.
u (psi)
u 1 = 31.2 Uavg
-
31.2 - (-51.2)
2
2
31.2 - 51.2
0 .
= -1 psi
2
a1
-
a2
---- -
------ =
41.2 psi
Ans.
These same results can also be obtained by drawing Mohr's
circle for each orientation of an element about the x, y, and z
axes, Fig.14-24d. Since a 1 and a 2 are of opposite signs, then the
absolute maximum shear stress as noted equals the maximum
in-plane shear stress.
14.5
ABSOLUTE MAXIMUM SHEAR STRESS
659
PROBLEMS
*14-76. Draw the three Mohr's circles that describe each of
the following states of stress.
14-78. Draw the three Mohr's circles that describe the
following state of stress.
25 ksi
5 ksi
~
(a)
(b)
Prob.14-78
Prob.14-76
14-79. Determine the principal stresses and the absolute
maximum shear stress.
14-77. Draw the three Mohr's circles that describe the
following state of stress.
'::!--.
30 psi
400 psi
Prob.14-77
Prob.14-79
660
CHAPTER
14
STRESS AND S TRA IN TRANSFORMATION
*14-80. Determine the principal stresses and the absolute
maximum shear stress.
14-82. Det ermine the principal stresses and the absolute
maximum shear stress.
5 ksi
Prob. 14-80
Prob. 14-82
14-81. Determine the principal stresses and the absolute
maximum shear stress.
14-83. Consider the general case of plane stress as shown.
Write a computer program that will show a plot of the three
Mohr's circles for the element. and will also determine the
maximum in-plane shear stress and the absolute maximum
shear stress.
Txy
u,
.
Prob. 14-81
Prob. 14-83
14.6
14.6
PLANE STRAIN
As outlined in Sec. 7.9, the general state of strain at a point in a body is
represented by a combination of three components of normal strain,
E.n E>" Ez, and three components of shear strain, 1'.tJ" l'xz, l'yz· The normal
strains cause a change in the volume of the element, and the shear
strains cause a change in its shape. Like stress, these six components
depend upon the orientation of the element, and in many situations,
engineers must transform the strains in order to obtain their values in
other directions.
To understand how this is done, we will direct our attention to a study
of plane strain , whereby the element is subjected to two components of
normal strain, Ex, Ey, and one component of shear strain, l'xy- Although
plane strain and plane stress each have three components lying in the
same plane, realize that plane stress does not necessarily cause plane
strain or vice versa. The reason for this has to do with the Poisson effect
discussed in Sec. 8.5. For example, the element in Fig. 14- 25 is subjected
to plane stress caused by <Tr and <Ty- Not only are normal strains Ex and Ey
produced, but there is also an associated normal strain, Ez, and so this is
not a case of plane strain.
Actually, a case of plane strain rarely occurs in practice, becall!se few
materials are constrained between rigid surfaces so as not to permit any
distortion in, say, the z direction. In spite of this, the analysis of plane
strain, as outlined in the following section, is still of great importance,
because it will allow us to convert strain-gage data, measured at a point
on the surface of a body, into plane stress at the point.
Plane stress, u,, u Y' does not cause plane
strain in the x- y plane since E, .,, 0.
Fig.14-25
PLANESTRAIN
661
662
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
GENERAL EQUATIONS OF
PLANE-STRAIN TRANSFORMATION
14.7
For plane-strain analysis it is important to establish strain transformation
equations that can be used to determine the components of normal and
shear strain at a point, Ex· , Ey'> 'Yx'y' , Fig. 14-26c, provided the components
Ex, Ey, 'Y.ry• are known, Fig. 14-26a. So in other words, if we know how the
element of material in Fig. 14-26a deforms, we want to know how the
tipped element of material in Fig. l4-26b will deform. To do this requires
relating the deformations and rotations of line segments which represent
the sides of differential elements that are parallel to the x,y and x ',y' axes.
Sign Convention. To begin, we must first establish a sign convention
for strain. The normal strains Ex and Ey in Fig. 14-26a are positive if they
cause elongation along the x and y axes, respectively, and the shear strain
'Yxy is positive if the interior angle AOB becomes smaller than 90°. This
sign convention also follows the corresponding one used for plane stress,
Fig. 14-Sa, that is, positive <Tn <Ty, 1:ry will cause the element to deform in
the positive Ex, Ey, 'Yxy directions, respectively. Finally, if the angle between
the x and x' axes is 8, then, like the case of plane stress, 8 will be positive
provided it follows the curl of the right-hand fingers, i.e., counterclockwise,
as shown in Fig. 14-26c.
__ ,
y
-
I
I
I
I
I
I
I
dy
I
I
--B
'"-".:....__r::.._--1.::....__
dx
x
11
(a)
y'
y
(c)
Fig.14-26
14.7
Normal and Shear Strains. To determine
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
we must find the
elongation of a line segment dx' that lies along the x' axis and is subjected
to strain components Ex, E,,, Yxr As shown in Fig. 14-27a, the components
of line dx' along the x and y axes are
dx
dy
= dx' cosB
= dx' sin 8
dx
Before deformation
y'
= E.r dx cos 0 + Ey dy sin 8 + Yxy dy cos 0
Since the normal strain along line dx' is
Eqs.14-15, we have
Ex·
y'
(14-15)
When the pos1t1ve normal strain Ex occurs, dx is elongated Ex dx,
Fig. 14-27b, which causes dx' to elongate Ex dx cos 8. Likewise, when Ey
occurs, dy elongates Ey dy, Fig. 14-27c, which causes dx' to elongate
Ey dy sin 0. Finally, assuming that dx remains fixed in position, the shear
strain Yxy in Fig. 14-27d, which is the change in angle between dx and dy,
causes the top of line dy to be displaced Yxy dy to the right. This causes
dx' to elongate Yxy dy cos 0. If all three of these (red) elongations are
added together, the resultant elongation of dx' is then
Bx'
y
Ex"
=
Ex
cos2 0 +
Ey
Ex· =
Normal strain
Bx' /dx', then using
sin2 8 + Yxy sin 8 cos 8
Ex
(b)
y
(14-16)
I
This normal strain is shown in Fig. 14-26b.
Normal strain
E
1
(c)
Shear strain Yxy
The rubber specimen is constrained between
lhe two fixed supports, and so it will undergo
plane strain when loads are applied to it in
lhe horizontal plane.
(d)
Fig. 14-27
663
664
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
y
y'
To determine 'Yx'y', we must find the rotation of each of the line
segments dx' and dy ' when they are subjected to the strain components
x'
/
Ex, Ey, 'Y.tJ'' Fust we will consider the counterclockwise rotation a of dx' ,
Fig. 14-27e. Here a = oy' /dx ' . The displacement oy' consists of three
I E_,dx cosO displacement components: one from e.n giving -ex dx sin(), Fig. 14-27b;
I o
another from Ey, giving Ey dy cos(), Fig. 14- 27c; and the last from 'Yxy•
giving - yX)' dy sin(), Fig.14- 27d. Thus, oy' is
( 1
I
I
dx
Normal strain E,
oy'
= -ex
dx sin() +
€ )'
dy cos() - 'Yxy dy sin()
(b)
y
Using Eq. 14-15, we therefore have
I
a,
Y _ (
a -_ dx'
- -ex
y'
+
)'a
a
Ey SIO v COS v -
· 2a
'Yxy SIO v
(14-17)
Finally, line dy' rotates by an amount {3, Fig. 14-27e. We can
determine this angle by a similar analysis, or by simply substituting
() + 90° for() into Eq. 14-17. Using the identities sin(() + 90°) = cos(),
cos(() + 90°) = -sin(), we have
Normal strain
{3 = (-ex
EY
= -(-ex
(c)
Y
y'
\
Yxydy sinO Yxydy
\
+ ey) sin(() + 90°) cos(() + 90°) -
;yxydy cosO
__ ::::::::_-,;;?"
/
+
Ey) cos() sin() - 'Yxy
sin2(e + 90°)
cos2 ()
Since a and {3 must represent the rotation of the sides dx' and dy' in
the manner shown in Fig. 14-27e, then the element is subjected to a shear
x' strain of
/'x'y' = a - {3 = -2(Ex - Ey)
Sin() COS() + 'Yx/cos2 ()
dx
Shear strain y xy
'Yxy
y
(d)
(e)
Fig. 14-27 (cont.)
-
sin2 ()) (14-18)
14.7
GENERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
y
y
y'
y'
Positive normal strain. Er•
Positive shear strain, Yxy
(a)
(b)
Fig. 14-28
Using
the
trigonometric
identities
sin 28 = 2 sine cos 8,
2
2
2
cos 8 = (1 + cos 28)/2, and sin 8 + cos 8 = 1, Eqs. 14- 16 and 14-18
can be written in the final form
Ex
Ex'=
'Yx'y'
2
+
Ey
+
2
= -
Ex -
Ey
2
(Ex -
Ey)
2
'Yxv
cos 28 +
- sin 28
2
'Yxy
sin28 + 2cos28
(14-19)
(14-20)
Normal and Shear Strain Components
According to our sign convention, if Ex· is positive, the element elongates
in the positive x' direction, Fig. l4-28a, and if 'Yx'y' is positive, the element
deforms as shown in Fig. 14-28b.
If the normal strain in the y' direction is required, it can be obtained
from Eq. 14-19 by simply substituting (8 + 90°) for 8. The result is
E •
y
=
E.r
+
2
Ey
-
Ex -
2
Ey
COS 28
~
/'.ry . ~
- -
2
Stll
(14-21)
The similarity between the above three equations and those for
plane-stress transformation, Eqs. 14-1, 14-2, and 14- 3, should be noted.
Making the comparison, <Tx, <Ty, O:r'> <Ty• correspond to E_., E,,, Ex·• e,,·; and
7:ry• 'Tx•y· correspond to 'Yxy/2, Y xy/2.
665
666
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
;
I
,
I
;
'
Principal Strains. Like stress, an element can be oriented at a point
so that the element's deformation is caused only by normal strains, with
no shear strain. When this occurs the normal strains are referred to as
principal strains, and if the material is isotropic, the axes along which
these strains occur will coincide with the axes of principal stress.
From the correspondence between stress and strain, then like
Eqs. 14-18 and 14-19, the direction of the x' axis and the two values of
the principal strains E 1 and E 2 are determined from
Complex stresses are often developed at
the joints where the cylindrical and
hemispherical vessels are joined together.
The stresses are determined by making
measurements of strain.
tan 28P
'Yxy
Ex - Ey
(14-22)
= ---
Orientation of principal planes
2 ('Yxy) 2
_
E
x
+ Ey )(Ex - Ey)
Et,2 +
+ 2
2
2
(14-23)
Principal strains
Maximum In-Plane Shear Strain. Similar to Eqs. 14-20, 14- 21,
and 14-22, the direction of the x' axis and the maximum in-plane shear
strain and associated average normal strain are determined from the
following equations:
tan 285
=
-(-E·_"_-_E-'y)
-'Yxy
(14-24)
Orientation of maximum in-plane shear strain
')'~;lane ~ (Ex ~ Ey)2 + (~ )2
1
(14-25)
Maximum in-plane shear strain
E avg =
2
(14-26)
Av·erage normal strain
IMPORTANT POINTS
• In the case of plane stress, plane-strain analysis may be used within the plane of the stresses to analyze
the data from strain gages. Remember, though, there will be a normal strain that is perpendicular to the
gages due to the Poisson effect.
• When the state of strain is represented by the principal strains, no shear strain will act on the element.
• When the state of strain is represented by the maximum in-plane shear strain, an associated average
normal strain will also act on the element.
14.7
EXAMPLE
14.12
The state of plane strain at a point has components of Ex = 500(10-6),
Ey = - 300(10-6)."Y.ry = 200(10-6), which tends to distort the element as
shown in Fig. 14-29a. Determine the equivalent strains acting on an
element of the material oriented clockwise 30".
SOLUTION
The strain transformation Eqs. 14-19 and 14-20 will be used to solve
the problem. Since 0 is positive counterclockwise, then for this problem
0 = - 30". Thus,
Ex·
=
Ex
+
Ey
2
667
GEN ERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
+
E.r -
Ey
2
Y
......---..~--~-
--1
I
~ ,:dy
I
1-·
/'xy .
cos 20 + 2sm 20
I
1
L
!
------
1tx
I
-1 :T?f
x
-1-1
£,<Lr
(a)
= [500 + ;-300))(l0-6) + [ 500 -i-300))(1o- 6)cos(2 (_30o))
y
y'
J
+ [ 200~0-6) sin(2(-30o))
= 213(10-6)
Ex•
"Yx'y'
2
Ex -
=-
(
= -[
Ey)
2
sin 20
"Yxy
+2
cos 20
500 - ( -300) ]
(10-6) sin(2(- 30"))
2
1'.r'y'
Ans.
+
200(10-6)
cos(2(- 30°))
2
= 793(10-6)
Ans.
(b)
The strain in the y' direction can be obtained from Eq. 14-21 with
0 = - 30". However, we can also obtain Ev· using Eq. 14-19 with
O = 60"(0 = -30° + 90°), Fig. 14-29b. We have with Ey• replacing Ex"
Ey•
=
Ex
+
2
Ey
+
Ex -
2
Ey
COS
20 +
/'xy
2
x'
y'
Sin 2()
= [500 + ;-300))(10- 6) + [500 -i-300))(10-6)cos(2(60o))
+
2
00(~0-6) sin(2(60°))
Ey•
= -13.4(10-6)
These results tend to distort the e lement as shown in Fig. 14-29c.
Ans.
(c)
Fig. 14-29
x'
668
I
14
CHAPTER
EXAMPLE
STRESS AND STRAIN TRANSFORMATI ON
14.13
The state of plane strain at a point has components of Ex = - 350(10-6),
6
Ey = 200(10-6), Y.tJ' = 80(10- ), Fig. 14-30a. Determine the principal
strains at the point and the orientation of the element upon which they act.
y
---;
SOLUTION
I ,__,/___/
Ey d y I
,- -
dv
,
- - -
I
I
Orientation of the Element. From Eq. 14- 22 we have
I
I
l--'-'-=-- -___,-------r
I
I
- :
----
1-
dx
tan28P =
y .Q'
2 --- x
--'--
y'
1-------. -=i-
E1dy'
-4.14° and 85.9°
Ans.
Principal Strains. The principal strains are determined from
Eq. 14-23. We have
_
EJ ,2 -
I
I
,
fd-----Li-~~~:·
C2dX
171.72°, so that
=
Each of these angles is measured positive counterclockwise, from the
x axis to the outward normals on each face of the element. The angle
of -4.l4°is shown in Fig. 14-30b.
I
I
I
85.9°
Ey
Thus, 28P = -8.28° and -8.28° + 180°
(;IP =
r-----.!.__I
Ex -
80(10- 6)
- ------( - 350 - 200)(10-6)
--~ E, dx
(a)
y
Yxy
---
=
Ex
+
2
Ey
2
E
y
)
2
+
f (Ex -
+ '\/
(Yxy )
2
-2-
(-350 + ~00)(10-6) + [ )~(-3-50 2-_-2_00_)~2_+_(_~0~
) 2 ](10-6)
1
(b)
=
Fig.14-30
-75.0(10- 6) + 277.9(10-6)
6
EJ = 203(10- )
Ans.
To determine the direction of each of these strains we will apply
Eq. 14-19 with 8 = -4.14°, Fig. 14-30b. Thus,
Ex
Ex• =
=
+
2
Ey
+
Ex -
2
Ey
'Yxy
COS
2(;1
+l
.
SJO
2fJ
(-350 2+ 200)(10-6) + (-350 2- 200)(10-6) cos 2(-4.140)
+
Ex• =
80(10- 6)
2
sin 2(-4.14°)
-353(10-6)
Hence Ex· = E 2. When subjected to the principal strains, the element is
distorted as shown in Fig. 14-30b.
14.7
EXAMPLE
14.14
The state of plane strain at a point has components of Ex = - 350(10-6),
"r = 200(10-6), Yxy = 80(10-6), Fig. 14-31a. Determine the maximum
in-plane shear strain at the point and the orientation of the element
upon which it acts.
y
SOLUTION
tan 2fJ
'
= -(
Ex -
Y xy
E1dy
From Eq. 14-24 we have
Orientation of the Element.
"r) = - -(- -350--- 200)(10-6)
----
I
l
80(10-6)
--
__ ....
I
I
I
I
---
y'
2
= [ )(-3502-
2
2
200) + (820) ] (10-6)
= 556(10- 6)
Y max
in-plane
Ans.
The square root gives two signs for y
max
• The
in-plane
proper one for each
angle can be obtained by applying Eq.14-20. When 8,
"Yx'y'
-- = -
Ex -
2
2
=
40.9°, we have
Ey sin 28 + -Yxy cos 28
2
= - ( - 35o - 200 )(10-6) sin 2(40.9°) +
2
"Y.r'y'
8
0(~0-6) cos 2(40.9°)
= 556(10-6)
This result is positive and so y
max
tends
in-plane
to distort the element so
that the right angle between dx' and dy' is decreased (positive sign
convention), Fig. 14-3lb.
Also, there are associated average normal strains imposed on the
element that are determined from Eq. 14-26.
=
Eavg
Ex+2
Ey
=
-350 + 200 ( 0_6)
2
1
= _
75 (l0-6)
These strains tend to cause the element to contract, Fig. 14-31b.
I
2
y
=)(Ex ~ Ey)2 + (~ )2
I
2 Y..,
(a)
Applying Eq. 14-25 gives
I
I
Notice that this orientation is 45° from that shown in Fig. 14-31b.
Maximum In-Plane Shear Strain.
---,
Y.o
dy
= 8 1.72° and 81.72° + 180° = 261.72°, so that
8s = 40.9° and 131°
Thus, 28,
669
GEN ERAL EQUATIONS OF PLANE-STRAIN TRANSFORMATION
(b)
Fag. 14-31
I
x
6 70
14
CHAPTER
STRESS AND STRAIN TRANSFORMATI ON
* 14. 8 MOHR'S CIRCLE-PLANE
STRAIN
Since the equations of plane-strain transformation are mathematically
similar to the equations of plane-stress transformation, we can also solve
problems involving the transformation of strain using Mohr's circle.
Like the case for stress, the parameter() in Eqs. 14-19 and 14-20 can be
eliminated and the result rewritten in the form
(Ex• -
2
€ avg)
+ ('Y
2'
xy
')2 - R
-
2
(14-27)
where
Equation 14-27 represents the equation of Mohr's circle for strain. It has a
center on the E axis at point C(Eavg• 0) and a radius R. As described in the
following procedure, Mohr's circle can be used to determine the principal
strains, the maximum in-plane strain, or the strains on an arbitrary plane.
PROCEDURE FOR ANALYSIS
The procedure for drawing Mohr's circle for strain follows
the same one established for stress.
Construction of the Circle.
Ex -
Ey
2
c 1--1
1----+----T------T---+-,~
Yxy
E.,+ Ey
Eavg =
2
_ _I
2 E ' ; - - ~1 9 =
')'
2
Fi.g .14-32
(f
E
• Establish a coordinate system such that the horizontal axis
represents the normal strain E, with positive to the right, and
the vertical axis represents half the value of the shear strain,
y /2, with positive downward, Fig. 14-32.
• Using the positive sign convention for Ex, Ey, 'Yx>" Fig. 14-26,
determine the center of the circle C, located Eavg = ( Ex + Ey)/2
from the origin, Fig. 14-32.
• Plot the reference point A having coordinates A(E.n 'Yxy/2). This
point represents the case when the x ' axis coincides with the
x axis. Hence() = 0°, Fig. 14-32.
• Connect point A with C and from the shaded triangle
determine the radius R of the circle, Fig. 14-32.
• Once R has been determined, sketch the circle.
Principal Strains.
14.8
6 71
MOHR'S CIRCLE-PLANE STRAIN
1 - - - - - E 1 - - - --1
• The principal strains E 1 and E2 are determined from the circle
as the coordinates of points B and D, that is, where y /2 = 0,
Fig. 14-33a.
• The orientation of the plane on which E 1 acts can be determined
from the circle by calculating 28p, using trigonometry. Here
this angle happens to be counterclockwise, measured from CA
to CB, Fig. 14-33a. Remember that the rotation of (}p, must be
in this same direction, from the element's reference axis x to
the x' axis, Fig.14- 33b.*
• When E 1 and E2 are positive as in Fig. 14-33a, the element in
Fig. 14-33b will elongate in the x' and y' directions as shown
by the dashed outline.
·1
F
D
B
1---~l°"'+--__:::'°'=~~--+.::_-- E
'Ymax
in- plane
, __ _ Ea\'g - - - 1
y
E
(a)
2
Maximum In-Plane Shear Strain.
• The average normal strain and half the maximum in-plane
shear strain are determined from the circle as the coordina~es
of point E or F, Fig.14-33a.
• The orientation of the plane on which y max
and E avg act can
in·plane
be determined from the circle in Fig. 14-33a, by calculating 285 ,
using trigonometry. Here this angle happens to be clockwise
from CA to CE. Remember that the rotation of (}51 must be in
this same direction, from the element's reference axis x to the
x' axis,Fig.14- 33c.*
Y
y ' Ymax
$__~-~~~ ---,
~~
I
I
I
Strains on Arbitrary Plane.
I
I
I
I
I
I
• The normal and shear strain components Ex· and 'Yx'y' for an
element oriented at an angle 8, Fig. 14-33d, can be obtained
from the circle using trigonometry to determine the coordina~es
of point P, Fig. 14- 33a.
• To locate P, the known counterclockwise angle(} of the x' axis,
Fig. 14- 33d, is measured counterclockwise on the circle as 28.
This measurement is made from CA to CP.
• If the value of Ey• is required, it can be determined by
calculating the E coordinate of point Qin Fig. 14-33a. The line
CQ lies 180° away from CP and thus represents a 90° rotation
of the x' axis.
I
I
I
I
.......~-;::;:-::::-~-:-IT-::;:::::::Tii--{.:f.=r:_o_,,__ x
/.-/
I
x'
'Ymax
in- plane
2
(c)
.Y'
)'
'Yxy -
2
-
,-
Eydy' L"
------
--·
I
I
I
I
I
l
l
I
I
I
I
I
I
'/xy
-
I
• _1..-.-/ 2
x'
yL~~-~-=-~-=-tj1E:=~10C::
i-l
x
Ex·dx'
*If the y /2 axis were constructed positive upwards, then the angle 20 on the circle
would be measured in the opposire direcrion to the orientation 0 of the plane.
(d)
Fig.14-33
672
14
CHAPTER
EXAMPLE
STRESS AND STRAIN TRANSFORMATI ON
14.15
-
.
The state of plane strain at a point has components of Ex = 250(10- 6),
6
Ey = -150(10- ) , 'Yxy = 120(10-6), Fig.14-34a. Determine the principal
strains and the orientation of the element upon which they act.
Y
-, -
'
/
I
I
,
f
1
dy
L /
,
SOLUTION
-- -- -·,,
- P'
'
----
--
1--
~ 'Yxy
2
+----'------ x
dx
--~,dx
Construction of the Circle. The E and y /2 axes are established in
Fig. 14- 34b. Remember that the positive y /2 axis must be directed
downward so that counterclockwise rotations of the element
. aroun d t he cuc
· 1e, an d vice
·
correspon d to countercf.ock wise rotation
versa. The center of the circle C is located at
(a)
=
250 + (-150)
(lo- 6 )
2
=
50(10- 6 )
Since Yxv /2 = 60(10- 6), the reference point A (8 = 0°) has
coordinates A(250(10-6), 60(10-6)). From the shaded triangle in
Fig. 14-34b, the radius of the circle is
R
= [
V(250 - 50)2 + (60) 2 ] (10- 6)
=
208.8(10-6)
Principal Strains. The E coordinates of points Band Dare therefore
-
250-
1. (10 - 6)
2
EJ
=
(50 + 208.8)(10- 6)
=
259(10- 6)
Ans.
E2
=
(50 - 208.8)(10- 6)
=
-159(10- 6)
Ans.
(b)
y'
y
I
The direction of the positive principal strain Et in Fig. 14- 34b is
defined by the counterclockwise angle 28p,, measured from
CA (8 = 0°) to CB. We have
_
-
_,..-
\€2dyc.l- - -------
--,
tan 28p, - -(-25_0_ _ _5_0_)
~
dy'
\
60
\
'
l,-
1--\-
I
dx' -
_ x'
-\-~--rj--:O-:p,---=8.35°
.I
'£ 1 d.~ '
(c)
Fig. 14-34
Ans.
X
Hence, the side dx' of the element is inclined counterclockwise 8.35°
as shown in Fig. 14- 34c. This also defines the direction of EJ . The
deformation of the element is also shown in the figure.
14.8
~ EXAMPLE
67 3
MOHR'S CIRCLE-PLANE STRAIN
14.16
The state of plane strain at a point has components of e_. = 250(10-6),
e,, = -150(10-6), 'Yxy = 120(10-6), Fig. 14-35a. Determine the
maximum in-plane shear strains and the orientation of the element
upon which they act.
Y
SOLUTION
The circle has been established in the previous example and is shown
in Fig. 14-35b.
Yxy
Maximum In-Plane Shear Strain. Half the maximum in-plane
shear strain and average normal strain are represented by the
coordinates of point E o r F o n the circle. From the coordinates of
point E,
2
(a)
('Y.r 'y ·)-·
1n,P1tn~
2
('Y.r 'Y•)ma.
in-pt11ne
=
418(10-6)
To orient the element, we will determine the clockwise angle 28,,,
measured from C4 (8 = O") to CE.
28,,
=
90° - 2(8.35°)
0,,
=
36.7°
Ans.
This angle is shown in Fig. 14-35c. Since the shear strain defined from
point E on the circle has a positive value and the average normal
strain is also positive, these strains deform the element into the dashed
shape shown in the figure.
y
y'
-
F
-1---11----=+<::--- - r l - - - ,f-E (10 - 6)
A
60
I
' o=oo
E{E 'Y~;.... \
~
ll\'& •
2
)
1--250--
i (l0 -6)
(c)
x'
(b)
Fig.14-35
674
-
14
CHAPTER
EXAMPLE
STRESS AND STRAIN TRANSFORMATI ON
14.17
The state of plane strain at a point has components of Ex = -300(10- 6),
6
6
Ey = -100(10- ) , 'Y.tJ' = 100(10- ) , Fig. 14-36a. Determine the state of
strain on an element oriented 20° clockwise from this position.
y
,'
, 'I
I
,'
I
j
Yxy
Z
dy
I
SOLUTION
Yx)'
2
I
The E and y /2 axes are established in
Fig. 14-36b. The center of the circle is at
Construction of the Circle.
x
dx
E.T dX
(a)
1 - - - -<E,· - - - - 1
The reference point A has coordinates A(-300(10- 6) , 50(10-6)), and
so the radius CA , determined from the shaded triangle, is
R
=
[
Y (300 -
200) 2 + (50) 2 ) (10-6)
=
111.8(10- 6)
Since the element is to be oriented
20° clockwise, we must consider the radial line CP, 2(20°) = 40°
clockwise, measured from CA (6 = 0°), Fig. 14- 36a. The coordinates
of point Pare obtained from the geometry of the circle. Note that
Strains on Inclined Element.
i -- -200- -- 1
1 - - - -300 - - - -1
1' (10 - 6)
2
(b)
y
y'
</> = tan- 1 ( (300
5~ 200)) =
Ex•
'Yx'y'
2
=
40° - 26.57°
=
-(200 + 111.8 COS 13.43°)(10-6)
=
-309(10-6)
=
-(111.8 sin 13.43°)(10- 6)
'Yx'y' =
Fig. 14-36
if!
=
13.43°
Thus,
I
(c)
26.570,
Ans.
-52.0(10- 6)
Ans.
The normal strain Ey• can be determined from the
point Q on the circle, Fig. 14-36b.
Ey• =
-(200 - 111.8 .cos 13.43°)(10-6)
=
E
coordinate of
-91.3(10-6)
Ans.
As a result of these strains, the element deforms relative to the x' , y '
axes as shown in Fig. 14-36c.
14.8
M OHR'S CIRCLE-PLANE STRAIN
675
PROBLEMS
*14-84. Prove that the sum of the normal strains in
perpendicular directions is constant. i.e., Ex + E,. = E.r + e,...
14-85. The state of strain at the point on the arm has
components of Ex = 200(1o-6), E1 = -300(10~). and
Yiy = 400(10~). Use the strain transformation equations to
determine the equivalent in-plane strains on an element
oriented at an angle of30" counterclockwise from the original
position. Sketch the deformed clement due to these strains
within the x- y plane.
*14-88. The state of strain at the point on the leaf of the
caster assembly has components of Ex = -400(10~) ,
e1 = 8()()(1o-6), and 1'xy = 375(lo-6). Use the strain
transformation equations to determine the equivalent
in-plane strains on an element oriented at an angle of
8 = 30" counterclockwise from the original position.
Sketch the deformed element due to these strains within
the x- y plane.
Prob. 14-88
Prob.14-85
14-86. The state of strain at the point on the pin leaf has
components of Ex = 200(10-"<i). Ey = 180(10~), and
.,,..,. = -300(10~). Use the strain transformation equations
and determine the equivalent in-plane strains on an element
oriented at an angle of 8 = 60" counterclockwise from the
original position. Sketch the deformed element due to these
strains within the x-y plane.
14-87. Solve Prob. 14-86 for an element oriented 8 = 30"
clockwise.
14-89. The state of strain at a point on the bracket has
component: Ex = 150(10"'6). Ey = 200(10"'6), y..,. = - 700(10"'6).
Use the strain transformation equations and determine the
equivalent in plane strains on an element oriented at an angle
of 8 = 60" counterclockwise from the original position. Sketch
the deformed element within the x-y plane due to these strains.
14-90. Solve Prob. 14-89 for an element oriented 8 = 30°
clockwise.
y
0 0
111----x
y
x
Probs. 14-86/87
Probs. 14-89/90
676
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
14-91. The state of strain at the point on the spanner
wrench has components of E, = 260(10-6), Ey = 320(10-6),
and 'Yxy = 180(10-6). Use the strain transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In
each case specify the orientation of the element and show
how the strains deform the element within the x-y plane.
14-93. The state of strain at the point on the support
has components of E, = 350(10-6), Ey = 400(10-6),
'Yxy = - 675{10-6). Use the strain transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
In each case specify the orientation of the element and show
how the strains deform the element within the x-y plane.
y
Prob.14-91
Prob.14-93
*14-92. The state of strain at the point on the member has
components of E., = 180(10-6), Ey = - 120(10-6), and
'Yxy = - 100(10-6). Use the strain transformation equations
to determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In
each case specify the orientation of the element and show
how the strains deform the element within the x-y plane.
Prob.14-92
14-94. Due to the load P, the state of strain at the
point on the bracket has components of E, = 500(10-6),
Ey = 350(10-6), and 'Yxy = - 430(10-6). Use the strain
transformation equations to determine the equivalent
in-plane strains on an element oriented at an angle of
e = 30° clockwise from the original position. Sketch the
deformed element due to these strains within the x-y plane.
Prob.14-94
14.8
14-95. The state of strain on an element has components
€ x = -400(10-6), Ey = 0, 'Yxy = 150(10-0). Determine the
equivalent state of strain on an element at the same point
oriented 30° clockwise with respect to the original element.
Sketch the results on this element.
*14-96.
The state of plane strain on the element is
6
£_, = - 300(10- ) , Ey = 0, and 'Y.<y = 150(10--<>). Determine
the equivalent state of strain which represents (a) the
principal strains, and (b) the maximum in-plane shear strain
and the associated average normal strain. Specify the
orientation of the corresponding elements for these states of
strain with respect to the original element.
y
T-
Consider the general case of plane strain where
and 'Yxy are known. Write a computer program that
can be used to determine the normal and shear strain, £ x'
and 'Yx'y', on the plane of an element oriented IJ from the
horizontal. Also, include the principal strains and the
element's orientation, and the maximum in-plane shear
strain, the average normal strain, and the element's
orientation.
14-98.
£,,Er,
14-99. The state of strain on the element has components
£, = -300(10-0),
Ey = 100(10-6),
'Yxy = 150(10-0).
Determine the equivalent state of strain, which represents
(a) the principal strains, and (b) the maximum in-plane
shear strain and the associated average normal strain.
Specify the orientation of the corresponding elements for
these states of strain with respect to the original element.
----
C!,
dy
677
MOHR'S CIRCLE-PLANE STRAIN
I
I
i-
I
I
J
)'
I
---'--i
I
-"----+---"=-~-+-'---- x
Yf
1--
~ ~i;,dx
-----
'YtY :
dy 2_
l
dx-l
Probs. 14-95196
1--
I
I
__ ..I--!
l!f/_j-E
2
14-97. The state of strain at the point on a boom of a shop
crane has components of £_, = 250(10--<>), £, = 300(10--<>),
'Yxy = - 180(10- 6 ). Use the strain transformation equations to
determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain. In
each case, specify the orientation of the element and show
how the strains deform the element within the x-y plane.
1--
x
x dx
dx
Prob.14-99
*14-100.
Solve Prob.14-86 using Mohr's circle.
14-101-
Solve Prob. 14-87 using Mohr's circle.
14-102-
Solve Prob. 14-88 using Mohr's circle.
14-103- Solve Prob. 14-91 using Mohr's circle.
Prob.14-97
*14-104. Solve Prob. 14-90 using Mohr's circle.
678
CHAPT ER
14
STR ESS AND STRAIN TRANSFORMATION
z
* 14.9 ABSOLUTE MAXIMUM SHEAR
STRAIN
In Sec. 14.5 it was pointed out that in the case of plane stress, the absolute
)'
x
x - y plane strain
maximum shear stress in an element of material will occur out ofthe plane
when the principal stresses have the same sign, i.e., both are tensile or both
are compressive. A similar result occurs for plane strain. For example, if
the principal in-plane strains cause elongations, Fig. 14- 37a, then the three
Mohr's circles describing the normal and shear strain components for the
element rotations about the x, y , and z axes are shown in Fig. 14-37b. By
inspection, the largest circle has a radius R = ( 'Yxz)max /2, and so
(a)
(14-28)
<' 1 and .-2
y
2
This value gives the absolute maximum shear strain for the material.
Note that it is larger than the maximum in-plane shear strain, which is
2
(y,,)max
2
( 'Yxy)max = Et -
E1.
Now consider the case where one of the in-plane principal strains is of
opposite sign to the other in-plane principal strain, so that Et causes
elongation and E 2 causes contraction, Fig. 14-38a. The three Mohr's
circles, which describe the strain components on the element rotated
about the x, y, z axes, are shown in Fig. 14-38b. Here
(b)
Fi.g .14-37
z
')'abs =
max
.- 1
)'
x
have the same sign
("'·)max
=Et
1 .r.y 1n·plane
-
E2
(14-29)
and .-2 have opposite signs
IMPORTANT POINTS
x - y plane strain
(a)
hxz)max
2
(yxy)max
y
2
2
(b)
Fi.g .14-38
• If the in-plane principal strains both have the same sign, the
absolute maximum shear strain will occur out of plane and has
a value of y abs = Emax· This value is greater than the maximum
.
max
.
m-plane shear stram.
• If the in-plane principal strains are of opposite signs, then the
absolute maximum shear strain equals the maximum in-plane
shear strain,')' abs = E t - E 2.
max
14.9
-
EXAMPLE 14.18
ABSOLUTE MAXIMUM SHEAR STRAIN
.
The state of plane strain at a point has strain components of
Ex = -400(10-6), Ey = 200(10-6), and 'Yxy = 150(10-6), Fig. 14-39a.
Determine the maximum in-plane shear strain and the absolute
maximum shear strain.
y
--+------ ---;
. I
~,dy ...L_
dy
l
II
I
I
I
/
I
Yx)'
-
I
2 Yx)':
2 I
I
400
-
'""-"-~----':...+..L---- x
1 -dx
(a)
(b)
Fig.14-39
SOLUTION
Maximum In-Plane Shear Strain. We will solve this problem using
Mohr's circle. The center of the circle is at
Eavg =
-400; 200 (10-6)
=
-100(10-6)
Since 'Yxy /2 = 75(10- 6 ), the reference point A has coordinates
(-400(10-6), 75(10- 6)), Fig.14-39b. The radius of the circle is therefore
R = [ V (400 - 100)2 + (75) 2 ] (10- 6) = 309(10- 6)
From the circle, the in-plane principal strains are
Et = (-100 + 309)(10- 6) = 209(10- 6)
6
E2 = (-100 - 309)(10- ) = -409(10-6)
Also, the maximum in-plane shear strain is
'Y!D•x
= Et - E 2 = (209 - (-409))(10- 6)
=
618(10- 6)
Ans.
1n·plane
Absolute Maximum Shear Strain. Since the principal in-plane
strains have opposite signs, the maximum in-plane shear strain is also
the absolute maximum shear strain; i.e.,
6
')' a bs = 618(10- )
max
Ans.
The three Mohr's circles, plotted for element orientations about each
of the x,y, z axes, are also shown in Fig. 14-39b.
679
680
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
14.10 STRAIN ROSETTES
cjj;I
'Ff
~'
The normal strain on the free surface of a body can be measured in a
particular direction using an e[ectrical resistance strain gage. For example,
in Sec. 8.1 we showed how this type of gage is used to find the axial strain
in a specimen when performing a tension test. When the body is subjected
to several loads, however, then the strains Ex, E>" l'xy at a point on its surface
may have to be determined. Unfortunately, the shear strain cannot be
directly measured with a strain gage, and so to obtain E.n Ey, l'xy• we must
use a cluster of three strain gages that are arranged in a specified pattern
called a strain rosette. Once these normal strains are measured, then the
data can be transformed to specify the state of strain at the point.
To show how this is done, consider the general case of arranging the
gages at the angles 8a, 8b, 8e shown in Fig. 14--40a. If the readings Ea, Eb, Ee
are taken, we can determine the strain components Ex, Ey, l'xy by applying
the strain transformation equation, Eq.14-16, for each gage. The results are
(a)
&:V
~
1 ~: 1
x
a
45• strain rosette
Ea = Ex
cos2 8a +
. 28
Ey sm
a
Eb = Ex
cos2 8 b +
Ey
. 28
sm
b +
l'.tJ' sin 8b
Ee = Ex
cos2 8c +
Ey
. 2 8
sm
c +
l'xy
(b)
\j.
~1§:1
x
+ l'xy sin 8a cos 8a
cos 8b
(14-30)
sin 8e cos 8c
The values of Ex, E>" l'.tJ' are determined by solving these three equations
simultaneously.
Normally, strain rosettes are arranged in 45° or 60° patterns. In the case
of the 45° or "rectangular" strain rosette, Fig. 14-40b, 8a = 0°, 8b = 45°,
8e = 90°, so that Eq. 14-30 gives
a
6(f strain rosette
(c)
Ey = Ee
Fig.14-40
-Y.t)' = 2Eb - ( E0
And for the 60° strain rosette, Fig. 14--40c, 8a
Here Eq. 14- 30 gives
Ey =
J'.t)' =
Typical electrical resistance 45°
strain rosette.
1
J
(2Eb
+ 2Ee
2
yj
( Eb -
=
-
+ Ee)
0°, 8b
Ea)
=
60°, 8e
=
120°.
(14-31)
Ee)
Once Ex, E>" l'xy are determined, then the strain transformation
equations or Mohr's circle can be used to determine the principal
or the maximum in-plane
shear strain I' 1n·p
!"""
•
in-plane strains E 1 and E1,
1ane
The stress in the material that causes these strains can then be determined
using Hooke's Jaw, which is discussed in the next section.
14.10
-
EXAMPLE
14.19
681
STRAIN ROSETIES
.
The state of strain at point A on the bracket in Fig.14-41a is measured
using the strain rosette shown in Fig. 14-41b. The readings from the
gages give Ea = 60(10- 6 ) , Eb = 135(10-6 ), and Ee = 264(10-6 ).
Determine the in-plane principal strains at the point and the
directions in which they act.
I
SOLUTION
We will use Eqs.14-30 for the solution. Establishing an x axis, Fig. 14-41b,
and measuring the angles counterclockwise from this axis to the
centerlines of each gage, we have 80 = 0°, 8b = 60°, and 8c = 120°.
Substituting these results, along with the problem data, into the equations
gives
60(10- 6 )
135(10- 6 )
= Ex
= ~
264(10-
E ),
sin2 0° + y xy sin 0° cos 0°
(1)
2
60° + Ey sin2 60° + 'Yxy sin 60° cos 60°
0.25Ex + 0.75Ey + 0.433/'.ry•
2
2
Ex cos 120° + Ey sin 120° + 'Y.ry•sin 120° cos 120°
0.25Ex + 0.75Ey - 0.433/'.ry•
= Ex cos
=
6
cos2 0° +
(a)
) =
=
x
(2)
a
(b)
(3)
Using Eq. 1 and solving Eqs. 2 and 3 simultaneously, we get
Ex
=
60(10- 6 )
Ey
246(10- 6 )
=
'Y.t)'
=
-149(10- 6 )
These same results can also be obtained in a more direct manner from
Eq.14-31.
The in-plane principal strains will be determined using Mohr's
circle. The center, C, is at Ea!f = 153(10- 6 ) , and the reference point
on the circle is at A(60(10- ), -74.5(10-6 )], Fig. 14-41c. From the
shaded triangle, the radius is
R
=
[v(153 - 60)
2
+ (74.5)
2
](lo-
6
)
=
(c)
119.1(10-6)
The in-plane principal strains are therefore
=
153(10-6) + 119.1(10-6)
=
272(10- 6 )
Ans.
E2 =
153(10-6) - 119.1(10-6)
=
33.9(10-6)
Ans.
Et
28p, = tan-
1
8Pz = 19.3°
(l 5~ ~ 60)
4
=
38.7°
Ans.
NOTE: The deformed element is shown in the dashed position in
Fig. 14-41d. Realize that, due to the Poisson effect, the element is
also subjected to an out-of-plane strain, i.e., in the z direction,
although this value will not influence the calculated results.
(d)
Fig. 14-41
682
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
14.11
MATERIAL PROPERTY
RELATIONSHIPS
In this section we will pres ent some important material property
relationships that are used when the material is subjected to multiaxial
stress and strain. In all cases, we will assume that the material 1s
homogeneous and isotropic, and behaves in a linear elastic manner.
Generalized Hooke's Law. When the material at a point is
subjected to a state of triaxial stress, a,,, ay, a z, Fig. 14-42a, then these
stresses can be related to the normal strains Ex, Ey, Ez by using the principle
of superposition, Poisson's ratio, e 13 1 = -ve100g, and Hooke's Jaw as it
applies in the uniaxial direction, e = a/ E. For example, consider
the normal strain of the element in the x direction, caused by separate
application of each normal stress. When a, is applied, Fig. l4-42b, the
element elongates with a strain e_~, where
a
e' = .....:!.
x
E
Application of ay causes the element to contract with a strain
Fig. l4-42c. Here
€"
x
=
ay
-v -
E
Finally, application of az, Fig. 14-42d, causes a contraction strain
so that
e"'
x
=
Uz
- v-
E
,
/
.........
I
I
I
I
I
I
'--' ;
,
,,,... "'
"'
(a)
•
.,,
: 1.........(,...,. :
~
I
I
I
+
,,
-
I
(T
y
iY"
I
I
~,,,-;.>-.._.._,
vz
"'
..,....,
... .._
I
...... ...
...
/
I
I
I
I
•,
.,
I
I
I
... ""-.)
,,"'
(c)
(b)
Fig.14-42
Ex,
"
(d)
e.~',
14.11
MATERIAL PROPERTY RELATIONSHIPS
We can obtain the resultant strain Ex by adding these three strains
algebraically. Similar equations can be developed for the normal strains
in they and z directions, and so the final results can be written as
Ex=
~fux -
v (u,.
+ uJ)
I
1
Ey = E fu,. - v(ux + uJ)
1
E;:
= E [u, - v(a:,
( 14-32)
+ uy)]
These three equations represent the general form of Hooke's law for a
triaxial state of stress. For application, tensile stress is considered a
positive quantity, and a compressive stress is negative. If a resulting
normal strain is positive, it indicates that the material elongates, whereas
a negative normal strain indicates the material contracts.
If we only apply a shear stress Txy to the element, Fig. 14-43a,
experimental observations indicate that the material will change its
shape, but it will not change its volume. In other words, Txy will only cause
the shear strain Y xy in the material. Likewise, Tyz and Tx z will only cause
shear strains Y yz and Y xz• Figs. 14-43b and 14-43c. Therefore, Hooke's law
for shear stress and shear strain becomes
1
= --r;
G Y",
Y y"..
(14-33)
I
'
'0
o
',,
1~
xy
.........
I
I
I
I
I
I
I
I
'~
I
I
I
I
I
.,,/ ! '
1
I
'
,
I
I
,
,_________ _
,,
I
I
I
(a)
(b)
(c)
Fi.g. 14-43
68 3
684
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
Relationship Involving E, v, and G. In Sec. 8.6 it was stated that
y
the modulus of elasticity Eis related to the shear modulus G by Eq. 8- 11,
namely,
(14-34)
l
1
'----~~~~~~~~~
x
(a)
y
Umin = - T:ry
One way to derive this relationship is to consider an element of the
material to be subjected only to shear, Fig. 14-44a. Applying Eq. 14-5
(see Example 14.5) the principal stresses at the point are umax = Try and
umin = -rry• where this element must be oriented ep, = 45°
counterclockwise from the x axis, as shown in Fig. 14-44b. If the three
principal stresses Umax = T")" uint = 0, and Umin = -r,y are then
substituted into the first of Eqs. 14-32, the principal strain Emax can be
related to the shear stress -i:ry · The result is
(14-35)
This strain, which deforms the element along the x' axis, can also be
related to the shear strain 'Yxy- From Fig. 14-44a, u, = u y = u z = 0.
Substituting these results into the first and second Eqs. 14- 32 gives
Ex = Ey = 0. Now apply the strain transformation Eq. 14-23, which gives
'--~~~~~~~~~ x
'Yxy
(b)
Et = €max=
T
Fi.g .14-44
By Hooke's law, 'Yxy = Try /G, so that Emax = -i:ry /2G. Finally, substituting
this into Eq. 14-35 and rearranging the terms gives our result, namely,
Eq.14-34.
dz
Dilatation. When an elastic material is subjected to normal stress, the
strains that are produced will cause its volume to change. For example, if
the volume element in Fig. 14-45a is subjected to the principal stresses
u1' u 2, u 3 , Fig. 14-45b, then the lengths of the sides of the element become
(1 + Ex) dx, (1 + Ey) dy, (1 + Ez ) dz. The change in volume of the
element is therefore
(a)
r
(1 + e, )dz
(1 + e,)dx
i
(b)
Expanding, and neglecting the products of the strains, since the strains
are very small, we get
The change in volume per unit volume is called the "volumetric strain"
or the dilatation e.
BV
x + €y + €z
e =-=E
Fi.g .14-45
dV
(14-36)
14.11
If we use Hooke's Jaw, Eq. 14-32, we can also express the dilatation in
u3 = p
terms of the applied stress. We have
Ie
=
1
~ 2v (
<Tt
+ <T2 +
<T3)
I
68 5
MATERIAL PROPERTY RELATIONSHIPS
j
(14-37)
Bulk Modulus. According to Pascal's Jaw, when a volume element of
material is subjected to a uniform pressure p caused by a static fluid , the
pressure will be the same in all directions. Shear stresses will not be
present, since the fluid does not flow around the element. This state of
" hydrostatic" loading therefore requires u 1 = u2 = u 3 = -p , Fig. 14-46.
Substituting into Eq. 14-37 and rearranging terms yields
p
-=
e
E
3(1 - 2v)
=
3(1 : 2v) ]
=p
Hydrostatic stress
(14-38)
The term on the right is called the volum e modulus of elasticity or the
bulk modulus, since this ratio,p / e,issimilar to the ratio of one-dimensional
linear elastic stress to strain, which defines E, i.e., <T / E = E. The bulk
modulus has the same units as stress and is symbolized by the Jetter k,
so that
[k
U2
UJ
(14-39)
For most metals v "" ~ and so k "" E. However, if we assume the
material did not change its volume when loaded, then av = e = 0, and k
would be infinite. As a result, Eq. 14-39 would then indicate the
theoretical maximum value for Poisson's ratio to be v = 0.5.
IMPORTANT POINTS
• When a homogeneous isotropic material is subjected to a state
of triaxial stress, the strain in each direction is influenced by
the strains produced by all the stresses. This is the result of the
Poisson effect, and the stress is then related to the strain in the
form of a generalized Hooke's Jaw.
• When a shear stress is applied to homogeneous isotropic
material, it will only produce shear strain in the same plane.
• The material constants£, G, and v are all related by Eq. 14-34.
• Dilatation, or volumetric strain, is caused only by normal strain,
not shear strain.
• The bulk modulus is a measure of the stiffness of a volume of
material. This material property provides an upper limit to
Poisson's ratio of v = 0.5.
Fig.14-46
=p
686
CHAPTER
14
EXAMPLE 14.20
STRESS AND STRAIN TRANSFORMATI ON
The bracket in Example 14.19, Fig. 14-47a, is made of steel for which
E,1 = 200 G Pa, v, 1 = 0.3. Determine the principal stresses at point A.
(a)
Fig. 14-47
SOLUTION I
From Example 14.19 the principal strains have been determined as
Et =
272(10-6)
€2
33.9(10-6)
=
Since point A is on the surface of the bracket, for which there is no loading,
the stress on the surface is zero, and so point A is subjected to plane stress
(not plane strain). Applying Hooke's Jaw with a 3 = 0, we have
272(10- 6)
=
54.4(106)
=
a1 - 0.3a2
33.9(10-6)
=
a2
200(109)
6.78(10 6)
=
a2 - 0.3a1
Ut
200(109)
-
-
03
·
a2
200(109)
(1)
0. 3 a 1
200(109)
(2)
Solving Eqs. 1 and 2 simultaneously yields
a 1 = 62.0MPa
Ans.
a 2 = 25.4MPa
Ans.
14.11
M ATERIAL PROPERTY RELATIONSHIPS
1---43.7 - --1
T
(MPa)
(b)
Fig. 14-47 (cont.)
SOLUTION II
It is also possible to solve this problem using the given state of strain
as specified in Example 14.19.
Ex
= 60( 10- 6)
= 246(10-6)
Ey
'Yxy
= -149(10-6)
Applying Hooke's law in the x- y plane, we have
60(10-6)
o:x =
=
<T.r
200(109) Pa
29.4 MPa
o:y =
200(109) Pa
0.30:,
200(109) Pa
58.0 MPa
The shear stress is determined using Hooke's law for shear. First,
however, we must calculate G.
E
G
= 2 (1 +
v)
=
200 GPa
2(1 + 0.3)
= 76.9 GPa
Thus,
1'.ry
=
Txy
G 'Yxy ;
= 76.9(109)(-149(10-6)] = -11.46 MPa
The Mohr's circle for this state of plane stress has a center at
o:avg = 43.7 MPa and a reference point A (29.4 MPa, -11.46 MPa),
Fig. 14-47b. The radius is determined from the shaded triangle.
R
= Y (43.7 -
29.4) 2 + (11.46) 2
= 18.3 MPa
Therefore,
= 43.7 MPa + 18.3 MPa = 62.0 MPa
u 2 = 43.7 MPa - 18.3 MPa = 25.4 MPa
<T1
Ans.
Ans.
NOTE: E ach of these solutions is valid provided the material is both
linear elastic and isotropic, since only then will the directions of the
principal stress and strain coincide.
68 7
688
I
CHAPTER
EX AMPLE
14
STRESS AND STRAIN TRANSFORMATI ON
14.21
The copper bar is subjected to a uniform loading shown in Fig.14-48. If it
has a length a = 300 mm, width b = 50 mm, and thickness t = 20 mm
before the load is applied, determine its new length, width, and thickness
after application of the load. Take Ecu = 120 GPa, v cu = 0.34.
SOOMPa
Fig.14-48
SOLUTION
By inspection, the bar is subjected to a state of plane stress. From the
loading we have
Ur = 800 MPa
Uy = -500 MPa
Try = 0,
Uz = 0
The associated normal strains are determined from Hooke's Jaw,
Eq. 14-32; that is,
u,
v
Ex = E - E (uy + u z)
Ey =
Ez =
4
800 MPa
0.3
(-500 MPa + 0)
120(103) MPa
120(103) MPa
Uy
V
E - E (ur + u z)
0 ·34
(800 MPa + 0)
- 500 MPa
3
3
120(10 ) MPa
120(10 ) MPa
Uz
V
E - E(ur + uy)
=
=
0.00808
-0.00643
0.34
(800 MPa - 500 MPa) = -0.000850
3
120(10 ) MPa
The new bar length, width, and thickness are therefore
=
0-
a' = 300 mm + 0.00808(300 mm) = 302.4 mm
Ans.
b ' = 50 mm + (-0.00643)(50 mm) = 49.68 mm
Ans.
t' = 20 mm + (-0.000850)(20 mm) = 19.98 mm
Ans.
14.11
I
EXAMPLE
MATERIAL PROPERTY RELATIONSHIPS
14.22
If the rectangular block shown in Fig. 14-49 is subjected to a uniform
pressure of p = 20 psi, determine the dilatation and the change in lenglh
of each side. Take E = 600 psi, v = 0.45.
Fig. 14-49
SOLUTION
Dilatation. The dilatation can be determined using Eq. 14-37 with
O:r = Uy = Uz = -20 psi. We have
e =
1 - 2v
E (Ur + Uy + u z)
=
1 - 2(0.45)
.
600 psi [3(-20 psi))
=
-0.01 in3/in3
Ans.
Change in Length. The normal strain on each side is detel!111ined
from Hooke's Jaw, Eq. 14-32; that is,
E
=
-
1
E [O:r - v(Uy
60
+ Uz)]
; psi [-20 psi - (0.45)( -20 psi - 20 psi)]
Thus, the change in length of each side is
oa = -0.00333(4 in.)
ob
=
=
-0.00333 in./ in.
-0.0133 in.
-0.00333(2 in.) = -0.00667 in.
oc = -0.00333(3 in.) = -0.0100 in.
The negative signs indicate that each dimension is decreased.
=
Ans.
Ans.
Ans.
689
690
CHAPTER
14
STRESS AND STRAIN TRANSFORMATI ON
PROBLEMS
14-105. The strain at point A on the bracket
has components E, = 300(10- 6 ) , E" = 550(10--<> ),
'Yxy = - 650(10--<> ), E, = 0. Determine (a) the principal
strains at A in the x-y plane, (b) the maximum shear strain
in the x-y plane, and (c) the absolute maximum shear strain.
14-107. The strain at point A on the pressure-vessel
wall has components E., = 480(10--<>), Ev = 720(10--<>),
'Yxy = 650(10--<>). Determine (a) the principaJ strains at A ,
in the x-y plane, (b) the maximum shear strain in the x-y
plane, and (c) the absolute maximum shear strain.
y
0
AL x
I'
y
A
x
'-
~
0
0
0
_Q.
0
0
0
0
0
0
0
0
0
o_
~
'
Prob.14-105
Prob.14-107
14-106. The strain at point A on a beam has components
Ex = 450(10- 6), Ey = 825(10- 6), 'Yxy = 275(10- 6), Ez = 0.
Determine (a) the principal strains at A , (b) the maximum
shear strain in the x-y plane, and (c) the absolute maximum
shear strain.
*14-108. The 45° strain rosette is mounted on the surface
of a shell. The following readings are obtained for each
- 200(10- 6), Eb 300(10- 6), and
gage: Ea Ee = 250(10- 6 ). Determine the in-plane principal strains.
Prob.14-106
Prob.14-108
14.11
14-109. For the case of plane stress, show that Hooke's
law can be written as
Ux
£
= --2-(Ex + llEy ).
(1 - v-)
Uy
=
£
? (Ey +
(l - v-)
llEx)
14-110. Use Hooke·s law. &j. 14-32. to develop the
strain tranformation equations, Eqs. 14-19 and 14-20, from
the stress tranformation equations, &js. 14-1 and 14-2.
M ATERIAL PROPERTY RELATIONSHIPS
691
14-115. The spherical pressure vessel has an inner
diameter of 2 m and a thickness of LO mm. A strain gage
having a length of 20 mm is attached to it, and it is observed
to increase in length by 0.012 mm when the vessel is
pressurized. Determine the pressure causing this
deformation, and find the maximum in-plane shear stress,
and the absolute maximum shear stress at a point on the
outer surface of the vessel. The material is steel, for which
E,. = 200 GPa and 1111 = 0.3.
14-111. The principal plane stresses and associated strains
in a plane at a point are u 1 = 36 ksi, u 2 = 16 ksi,
3
Et = 1.02(10-3), Ez = 0.180(10- ). Determine the modulus
of elasticity and Poisson's ra tio.
*14-112. A rod has a radius of 10 mm. If it is subjected to
an axial load of 15 N such that the axia l strain in the rod is
Ex = 2.75(10-6), determine the modulus of elasticity E and
the change in the rod's diameter. 11 = 0.23.
Prob. 14-115
14-113. The polyvinyl chloride bar is subjected to an axial
force of 900 lb. If it has the original dimensions shown,
determine the change in the angle 8 after the load is applied.
£P'°< = 800(103) psi. "P'"< = Q.20.
14-114. The polyvinyl chloride bar is subjected to an
axial force of 900 lb. If it has the original dimensions
shown, determine the value of Poisson's ratio if the angle 8
decreases by ti.8 = 0.0 I 0 after the load is applied.
Epvc = 800(103) psi.
900lb--=r=
3in.
*14-116. Determine the bulk modulus for each of the
following materials: (a) rubber.£, = 0.4 ksi, 11, = 0.48, and
(b) glass, Eg = 8(la3) ksi. vg = 0.24.
14-117. The strain gage is placed on the surface of the steel
boiler as shown. If it is 0.5 in. long. determine the pressure in
the boiler when the gage elongates 0.2(10-3) in. The boiler
has a thickness of 0.5 in. and inner diameter of 60 in. Also,
determine the maximum x, y in-plane shear strain in the
material.£,. = 29(103) ksi, 11., = 0.3.
_____., 900 1b
--.1...._6-in-.-=-====~R_ I in.
_l._ ""
11-r::;,.-----
Probs. 14-113/L14
Prob. 14-117
692
CHAPT ER
14
STR ESS AND STRAIN TRANSFORMATION
14-118. The principal strains at a point on the alum inum
fuselage of a jet aircraft are £ 1 = 780(10- 6) and
£ 2 = 400(10- 6). Determine the associated principal stresses
at the point in the same plane. Ea1 = lO(lo-1) ksi, va1 = 0.33.
Hinr: See Prob. 14-109.
14-122. The principal strains at a point on the aluminum
surface of a tank are £ 1 = 630(10- 6) and £ 2 = 350(10-6). If
this is a case of plane stress, determine the associated
principal stresses at the point in the same plane.
Ea1 = 10(103) ksi, v31 = 0.33. Hinr: See Prob. 14-109.
14-119. The strain in the x direction at point A on the
A-36 structural-steel beam is measured and found to be
£., = 200(10-6). Determine the applied load P. What is the
shear strain Yxy at point A?
14-123. A uniform edge load of 500 lb/in. and 350 lb/in.
is applied to the polystyrene specimen. If the specimen is
originally square and has dimensions of a = 2 in., b = 2 in.,
and a thickness of 1 = 0.25 in., determine its new dimensions
a', b' , and r' after the load is applied. EP = 597(103) psi and
Vp = 0.25.
*14-120. If a load of P = 3 kip is applied to the A-36
structural-steel beam, determine the strain Ex and Yxy at
point A.
I
t - - 3 ft - -+ - - - - 4 ft
-1
I
8
500 lb/in.
a= 2 in
p
!
'
r
y
[2 in.
350 lb/in.
l_o_
lfl
2 in.
l
-
_J12in.
-116 in.
l - b=2in. - J
Prob.14-U3
Probs. 14-119/UO
*14-124. A material is subjected to principal stresses u., and
u >" Determine the orientation 8 of the strain gage so that its
reading of normal strain responds only to u Y and not u ,. The
material constants are E and v.
14-Ul. The cube of aluminum is subjected to the three
stresses shown. Determine the principal strains. Take
Ea1 = lO(lfrl) ksi and va1 = 0.33.
y
26 ksi
15 ksi
Prob.14-Ul
Prob.14-U4
693
CHAPTER REVIEW
CHAPTER REVIEW
y
Plane stress occurs when the material at a point
is subjected to two normal stress components
u,, and u,. and a shear stress T,,,.. Provided these
components are known. then the stress
components acting on an element having a
different orientation 8 can be determined using
tbe two force equations of equilibrium or the
equations of stress transformation.
Ux·
Tx•y•
=
Ux
+ Uy
2
=-
+
Ux - Uy
2
Ux - u,,
2
sin 28 +
COS
28 +
Txy
cos 28
Txy
I
Uy
T xy
x
Ux
sin 28
x'
0
For design. it is important to determine the
orientation of the element that produces
the maximum principal normal stresses and the
maximum in-plane shear stress. Using the stress
transformation equations, it is found that no
shear stress acts on the planes of principal
stress. The principal stresses are
Ut.2
ux +2 u,. ± J(u,, -2
=
Uy)2
+
2
T xy
Ux
-
Txy
<T.,.,
The planes of maximum in-plane shear stress
are oriented 45° from this orientation, and on
these shear planes there is an associated
average normal stress.
u,) +
2
in·planc
2
Tm:uc
••·1•''"'
O'avg
= J(Ux -
=
U.r
+ Uy
2
T max
T
2
xy
694
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
Mohr's circle provides a semi-graphical
method for finding the stress on any
plane, the principal normal stresses,
and the maximum in-plane shear stress.
To draw the circle, the u and Taxes are
established, the center of the circle
C((ux + u1,)/2, OJ and the reference
point A(u" Txy ) are plotted. The radius
R of the circle extends between these
two points and is determined from
trigonometry.
Ux-------l
R=
J(""-"r)2
2
If u 1 and u 2 are of the same sign, then
the absolute maximum shear stress
will lie out of plane.
In the case of plane stress, the absolute
maximum shear stress will be equal to
the maximum in-plane shear stress
provided the principal stresses u 1 and
u 2 have the opposite sign.
x- y plane stress
x - y plane stress
UJ -
Tabs max
When an element of material is
subjected to deformations that only
occur in a single plane, then it
undergoes plane strain. If the strain
components"·" ">" and Yxy are known
for a specified orientation of the
element, then the strains acting for
some other orientation of the
element can be determined using the
plane-strain transformation equations.
Likewise, the principal normal strains
and maximum in-plane shear strain can
be determined using transformation
equations.
Tabs -
2
ma.-.:
Ex+ Ey
2
Ex• =
E.t -
+
2
E.t -
2
°Y<'y'
~
EJ.2
Ey
Ey
'Yxv
cos 28 + 2sin 28
'Yxv
cos 28 - 2sin 28
"Yxv
= - ("'· -2 "Y)sin28 + 2cos28
=
E.t
+
2
E)'
+
2
U2
+ -r,;
695
CHAPTER REVlEW
Strain transformation problems can
also be solved in a semi-graphical
manner using Mohr's circle. To draw
the circle, the E and y /2 axes are
established and the center of the
circle C [(Ex + E1)/2. OJ and the
""reference point" A (Ex. Yxy/2) are
ploned. The radius of the circle
extends between these two points
and is determined from trigonometry.
1----+---C"*I---;I-;- . - E
1----,.--1----~~
I
Ea'"I
y
2
If the material is subjected to
triaxial stress, the n the strain 111
each direction is innuenced by the
strain produced by all three stresses.
Hooke's law then involves the
material properties £and 11.
1
= - [u. - v(ux + u1)]
•
E '
E.
If £ and 11 are known. then G can
be determined.
G=
E
2(1 + 11)
1 - 211
The dilatation is a measure of
volume tric strain.
e=
The bulk modu lus is used to
measure the stiffness of a volwne
of material.
E
k= - - - 3(1 - 211)
£
(o:, + u1 + u,)
~ :___]
2 E-:---/,,. , .11 0 =Cf'
E,+ E,
=
y~
696
CHAPTER
14
STRESS AND STRAIN TRANSFORMATION
REVIEW PROBLEMS
R14-1. The steel pipe has an inner diameter of 2.75 in.
and an outer diameter of 3 in. If it is fixed at C and subjected
to the horizontal 20-lb force acting on the handle of the pipe
wrench, determine the principal stresses in the pipe at
point A, which is located on the surface of the pipe.
R14-3. Determine the equivalent state of stress if an
element is oriented 40° clockwise from the element shown.
Use Mohr's circle.
201b
r
L
110 ksi
10 in.
>---•- 6ksi
B
c
y
Prob. R14-3
Prob. R14-1
R14-2. The steel pipe has an inner diameter of 2.75 in.
and an outer diameter of 3 in. If it is fixed at C and subjected
to the horizontal 20-lb force acting on the handle of the pipe
wrench, determine the principal stresses in the pipe at
point B, which is located on the surface of the pipe.
*R14-4. The crane is used to support the 350-lb load.
Determine the principal stresses acting in the boom at
points A and B. The cross section is rectangular and has a
width of 6 in. and a thickness of 3 in. Use Mohr's circle.
201b
r
10 in.
L
B
c
y
Prob. R14-2
Prob. R14-4
REVIEW PROBLEMS
Rl4-5. In the case or plane stress, where the in-plane
principal strains are given by E1 and E2 , show that the third
principal strain can be obtained from
697
*Rl4-8. A single strain gage. placed in the vertical plane
on the outer surface and at an angle 60° to the axis of the
pipe, gives a reading at point A of EA = -250(1~).
Determine the principal strains in the pipe at this point. The
pipe bas an outer diameter or 1 in. and an inner diameter of
0.6 in. and is made of C86100 bronze.
where vis Poisson·s ratio for the material.
R14-6. The plate is made of material having a modulus of
elasticity£= 200 GPa and Poisson·s ratio v = Determine
the change in width a, height b. and thickness c when it is
subjected to the uniform distributed loading shown.
i.
2MN/m
Prob. Rl4-8
I
y
b=300mm
j
x
:z:
Prob. Rl4-6
Rl4-9. The 60° strain rosette is mounted on a beam. The
following readings arc obtained for each gage:
E0 = 600(10....,), Eb= -700(10-<>), and Ee = 350(10....,).
Determine (a) the in-plane principal strains and (b) the
maximum in-plane shear strain and average normal strain.
1n each case show the deformed clement due to these
strains.
Rl4-7. If the material is graphite for which £, = 800 ksi
and = 0.23, determine the principal strains.
v,
26 ksi
c
15 ksi
Prob. Rl4-7
Prob. Rl4-9
I
CHAPTER
15
(©Olaf Speier/Alamy)
Beams are important structural members used to support roof and floor loadings.
DESIGN OF BEAMS
AND SHAFTS
CHAPTER OBJECTIVES
•
To develop methods for designing beams to resist both bending
and shear loads.
15.1
BASIS FOR BEAM DESIGN
Beams are said to be designed on the basis ofstrength when they can resist
the internal shear and moment developed along their length. To design a
beam in this way requires application of the shear and flexure formulas
provided the material is homogeneous and has linear elastic behavior.
Although some beams may also be subjected to an axial force, the effects
of this force are often neglected in design since the axial stress is generally
much smaller than the stress developed by shear and bending.
699
700
CHAPTER
15
DES I GN OF BEAMS AND SHAFTS
As shown in Fig. 15- 1, the external loadings on a beam will create
additional stresses in the beam directly under the load. Notably, a
compressive stress a y will be developed, in addition to the bending stress
a, and shear stress 7:ry discussed previously in Chapters 11 and 12. Using
advanced methods of analysis, as treated in the theory of elasticity, it can
be shown that a y diminishes rapidly throughout the beam's depth, and
for most beam span-to-depth ratios used in engineering practice, the
maximum value of ay remains small compared to the bending stress a .,
that is, a, >> a y- Furthermore, the direct application of concentrated
loads is generally avoided in beam design. Instead, bearing plates are
used to spread these loads more evenly onto the surface of the beam,
thereby further reducing a y.
Beams must also be braced properly along their sides so that they do
not sidesway or suddenly become unstable. In some cases they must also
be designed to resist deflection, as when they support ceilings made of
brittle materials such as plaster. Methods for finding beam deflections
will be discussed in Chapter 16, and limitations placed on beam sidesway
are often discussed in codes related to structural or mechanical design.
Knowing how the magnitude and direction of the principal stress
change from point to point within a beam is important if the beam is
made of a brittle material, because brittle materials, such as concrete, fail
in tension. To give some idea as to how to determine this variation, Jet's
consider the cantilever beam shown in Fig. 15- 2a, which has a rectangular
cross section and supports a load P at its end.
In general, at an arbitrary section a- a along the beam, Fig. 15- 2b, the
internal shear V and moment M create a parabolic shear-stress
distribution and a linear normal-stress distribution, Fig. 15- 2c. As a result,
the stresses acting on elements located at points 1 through 5 along the
section are shown in Fig. 15- 2d. Note that elements 1 and 5 are subjected
only to a maximum normal stress, whereas element 3, which is on the
neutral axis, is subjected only to a maximum in-plane shear stress. The
intermediate elements 2 and 4 must resist both normal and shear stress.
Uy
y
p
+
+++u,
'l"xy
!1 rm-ITT1
D
~
D
u.
+++u,
+'xy
Fig.15-1
x
15.1
BASIS FOR BEAM D ESIGN
p
a
When these states of stress are transformed into principal s1resses,
using either the stress transformation equations or Mohr's circle, the
results will look like those shown in Fig. 15- 2e. If this analysis is extended
to many vertical sections along the beam other than a-a, a profile of the
results can be represented by curves called stress trajectories. Each of
these curves indicates the direction of a principal stress having a constant
magnitude. Some of these trajectories are shown in Fig. 15- 3. Here the
solid lines represent the direction of the tensile principal stresses and the
dashed lines represent the direction of the compressive principal stresses.
As expected, the lines intersect the neutral axis at 45° angles
(like element 3), and the solid and dashed lines will intersect at 90°
because the principal stresses are always 90° apart. Once the directions
of these lines are established, it can help engineers decide where and
how to place reinforcement in a beam if it is made of brittle mate rial, so
that it does not fa il.
Stress trajectories for
cantilevered beam
Fig. 15-3
701
702
CHAPTER
15
DES I GN OF BEAMS AND SHAFTS
15.2
PRISMATIC BEAM DESIGN
Most beams are made of ductile materials, and when this is the case it is
generally not necessary to plot the stress trajectories for the beam. Instead,
it is simply necessary to be swre the actual bending and shear stress in the
beam do not exceed allowable limits as defined by structural or mechanical
codes. In the majority of cases the suspended span of the beam will be
relatively long, so that the internal moments within it will be large. When
this is the case, the design is then based upon bending, and afterwards the
shear strength is checked.
A bending design requires a determination of the beam's section
modulus, a geometric property which is the ratio of I to c, that is, S = I/ c.
Using the flexure formula, a = Mc/ I, we have
Sreq'd
A
B
The two floor beams are connected to the
beam AB, which transmits the load to the
columns of this building frame. For design,
all the connections can be considered to act
as pins.
Mmax
= --
(15- 1)
Uallow
Here Mmax is determined from the beam's moment diagram, and the
allowable bending stress, aauow• is specified in a design code. In many cases
the beam's as yet unknown weight will be small, and can be neglected in
comparison with the loads the beam must carry. However, if the additional
moment caused by the weight is to be included in the design, a selection
for s is made so that it slightly exceeds sreq'd·
Once Sreq'd is known, if the beam has a simple cross-sectional shape,
such as a square, a circle, or a rectangle of known width-to-height
proportions, its dimensions can be determined directly from Sreq'd, since
Sreq"d = I/c. However, if the cross section is made from several elements,
such as a wide-flange section, then an infinite number of web and flange
dimensions can be determined that satisfy the value of Sreq'd· In practice,
however, engineers choose a particular beam meeting the requirement
that S > S,eq'd from a table that lists the standard sizes available from
manufacturers. Often several beams that have the same section modulus
can be selected, and if deflections are not restricted, usually the beam
having the smallest cross-sectional area is chosen, since it is made of Jess
material, and is therefore both lighter and more economical than
the others.
15.2
PRISMATIC
BEAM
703
DESIGN
Once the beam has been selected, the shear formula can then be used
to be sure the allowable shear stress is not exceeded, TaJJow > VQ/ It.
Often this requirement will not present a problem; however, if the beam
is "short" and supports large concentrated loads, the shear-stress
limitation may dictate the size of the beam.
Steel Sect.
Most manufactured steel beams are produced by
rolling a hot ingot of steel until the desired shape is formed. These
so-called rolled shapes have properties that are tabulated in the
American Institute of Steel Construction (AISC) manual. A
representative listing of different cross sections taken from this manual is
given in Appendix B. Here the wide-flange shapes are designated by
their depth and weight per unit length; for example, W18 x 46 indicates
a wide-Clange cross section (W) having a depth of 18 in. and a weight of
46 lb/ft, Fig. 15-4. For any given selection, the weight per length,
dimensions, cross-sectional area, moment of inertia, and section modulus
are reported. Also included is the radius of gyration, r, which is a
geometric property related to the section's buckling strength. This will be
discussed in Chapter 17.
Typical profile view of a steel
wide-flange beam
0.605 in.
T __
18 in.
0.360 in.
1,....---' . ____,
The large shear force that occurs at the
support of this steel beam can cause
localized buckling of the beam's Oanges
or web. To avoid this, a "'stiffener" A is
placed along the web to maintain stability.
Wood Sect o
Most beams made of wood have rectangular cross
sections because such beams are easy to manufacture and handle.
Manuals, such as that of the National Forest Products Association, List
the dimensions of lumber often used in the design of wood beams. Lumber
is identified by its nominal dimensions, such as 2 x 4 (2 in. by 4 in.);
however, its acwal or "dressed" dimensions are smaller, being 1.5 in.
by 3.5 in. The reduction in the dimensions occurs in order to obtain a
smooth surface from lumber that is rough sawn. Obviously, the acwal
dimensions must be used whenever stress calculations are performed on
wood beams.
l-6in.- J
W18 X 46
Fig. 15-4
704
CHAPTER
15
DES I GN OF BEAMS AND SHAFTS
'
Welded
Bolted
Steel plate girders
Fig.15-5
Built-up Sections. A built-up section is constructed from two or
Wooden box beam
(a)
more parts joined together to form a single unit. The capacity of this
section to resist a moment will vary directly with its section modulus, since
sreq'd = MI <Tallow· If sreq'd is increased, then so is I because by definition
Sreq'd = I/ c. For this reason, most of the material for a built-up section
should be placed as far away from the neutral axis as practical. This, of
course, is what makes a deep wide-flange beam so efficient in resisting a
moment. For a very large load, however, an available rolled-steel section
may not have a section modulus great enough to support the load. When
this is the case, engineers will usually "build up" a beam made from plates
and angles. A deep I-shaped section having this form is called a plate
girder. For example, the steel plate girder in Fig.15- 5 has two flange plates
that are either welded or, using angles, bolted to the web plate.
Wood beams are also "buillt up," usually in the form of a box beam,
Fig. 15-6a. They may also be made having plywood webs and larger
boards for the flanges. For very large spans, glulam beams are used.
These members are made from several boards glue-laminated together
to form a single unit, Fig. 15-6b.
Just as in the case of rolled sections or beams made from a single piece,
the design of built-up sections requires that the bending and shear stresses
be checked. In addition, the shear stress in the fasteners, such as weld, glue,
nails, etc., must be checked to be certain the beam performs as a single unit.
IMPORTANT POINTS
Glulam beam
(b)
Fig.15-6
• Beams support loadings that are applied perpendicular to
their axes. If they are designed on the basis of strength, they
must resist their allowable shear and bending stresses.
• The maximum bending stress in the beam is assumed to be much
greater than the localized stresses caused by the application of
loadings on the surface of the beam.
15.2
PRISMATIC BEAM DESIGN
PROCEDURE FOR ANALYSIS
Based on the previous discussion, the following procedure provides a rational method for the design of a
beam on the basis of strength.
Shear and Moment Diagrams.
• Determine the maximum shear and moment in the beam. Often this is done by constructing the
beam's shear and moment diagrams.
Bending Stress.
• If the beam is relatively long, it is designed by finding its section modulus using the flexure formula,
Sreq·d =
Mmax/aauow·
• Once Sreq'd is determined, the cross-sectional dimensions for simple shapes can then be calculated,
since Sreq·d = I/c.
• If rolled-steel sections are to be used, several possible beams can be selected from the tables in
Appendix B that meet the requirement that S > Sreq'd· Of these, choose the one having the smallest
cross-sectional area, since this beam has the least weight and is therefore the most economical.
• Make sure that the selected section modulus, S, is slightly greater than Sreq'd• so that the additional
moment created by the beam's weight is considered.
Shear Stress.
• Normally beams that are short and carry large loads, especially those made of wood, are first designed
to resist shear and then later checked against the allowable !bending stress requirement.
• Using the shear formula, check to see that the allowable shear stress is not exceeded; that is, use
Tallow
> Ymax Q/ ft.
• If the beam has a solid rectangular cross section, the sheair formula becomes Tallow <:: 1.5 (Vmax/ A)
(see Eq. 2 of Example 12.2.), and if the cross section is a wide flange, it is generally appropriate to
assume that the shear stress is constant over the cross-sectional area of the beam's web so that
Tallow
> Vmax/ Aweb• where A.veb is determined from the product of the web's depth and its thickness.
(See the note at the end of Example 12.3.)
Adequacy of Fasteners.
• The adequacy of fasteners used on built-up beams depends upon the shear stress the fasteners can
resist. Specifically, the required spacing of nails or bolts of a particular size is determined from the
allowable shear flow, qallow = VQ/ I, calculated at points on the cross section where the fasteners are
located. (See Sec. 12.3.)
705
706
I
15
CHAPTER
EXAMPLE
DESIGN OF BEAMS AND SHAFTS
15.1
40 kip
20 kip
1
K
l -6ft -1-
,;
i
6ft -1· - 6ft -1
A beam is to be made of steel that has an allowable bending stress of
<Tallow = 24 ksi and an allowable shear stress of Tallow = 14.5 ksi. Select an
appropriate W shape that will carry the loading shown in Fig. 15-7a.
SOLUTION
The support reactions have been
calculated, and the shear and moment diagrams are shown in Fig. 15-7b.
From these diagrams, Vmax = 30 kip and Mmax = 120 kip· ft.
Shear and Moment Diagrams.
(a)
40 kip
20kip
'
i
The required section modulus for the beam ts
determined from the flexure formula ,
Bending Stress.
S
•
-. - 6 ft -
-
6 ft - t -6 ft -
10 kip
50 kip
'd =
req
20
I
W18
W16
W14
W12
WlO
W8
x (ft)
_"0
_, 1I'-_ ___,_
M (kip·ft)
<Tallow
=
120 kip· ft (12 in./ft)
. 3
=
60m
24 k'tp /'Ill2
Using the table in Appendix B, the following beams are adequate:
I
~:rp)
Mmax
x
x
x
x
x
x
40
45
43
50
54
67
S = 68.4 in3
S
S
S
S
S
=
=
=
=
=
72.7 in3
62.7 in3
64.7 in3
60.0 in3
60.4 in3
60
The beam having the least weight per foot is chosen,* i.e.,
I""'-- - -->,.- - ' - - - -....;..- x (ft)
W18 x 40
Shear Stress. Since the beam is a wide-flange section, the average
shear stress within the web will be considered. (See Example 12.3.) Here
- 120
(b)
the web is assumed to extend from the very top to the very bottom of the
beam. From Appendix B, for a W18 x 40, d = 17.90 in., f w = 0.315 in.
Thus,
Fig.15-7
'Tavg
vmax
30 kip
= _A_w_ = (l _ in.)(0.3l in.) = 5.32 ksi
Use a W18 x 40.
7 90
5
< 14.5 ksi OK
Ans.
*The additional moment caused by the weight of the beam, (0.040 kip/ft) (18 ft}= 0.720 kip,
will only slightly increase Srcq'd ·
15.2
I
EXAMPLE
1 s.2
707
PRISMATIC BEAM DESIGN
j
The laminated wooden beam shown in Fig. 15--8a supports a uniform
distributed loading of 12 kN /m. If the beam is to have a height-to-width
ratio of 1.5, determine its smallest width. Take aauow = 9 MPa, and
Tallow = 0.6 MPa. Neglect the weight of the beam.
SOLUTION
(a)
The support reactions at A and B
have been calculated, and the shear and moment diagrams are shown in
Fig. 15--8b. Here Vmax = 20 kN, Mmax = 10.67 kN · m.
Shear and Moment Diagrams.
Applying the flexure formula,
Bending Stress.
_
Mmax _
sreq·d -
-
aauow
10.67(103) N · m _
6
2
9(10 ) N /m
-
32kN
3
0.00119 m
V(kN)
20
Assuming that the width is a, then the height is l.5a, Fig. 15- 8a. Thus,
s.
req d
I
=-= 000119
c
.
(1 5a)3
1.(a)
12
=
.
m
(0.75a)
3
- 12
a 3 = 0.003160 m3
a = 0.147 m
- 16
M(kN·m)
10.67
Applying the shear formula for rectangular sections
(which is a special case of Tmax = VQ/ It, as shown in Example 12.2),
we have
Shear Stress.
Vmax
Tmax
= l. 5 A
-6
20(1a3) N
5
= (1. ) (0.147 m)(l.5)(0.147 m)
Fig.15-8
= 0.929 MPa > 0.6 MPa
Since the design based on bending fails the shear criterion, the beam must
be redesigned on the basis of shear.
Tallow =
1.5
Vmax
A
20(103) N
600 kN/m - 1.5 (a)(l. 5a)
2
_
a = 0.183 m = 183 mm
This larger section will also adequately resist the bending stress.
(b)
Ans.
708
I
CHAPTER
15
DES I GN OF BEAMS AND SHAFTS
15.3 1
EXAMPLE
The wooden T-beam shown in Fig. l5- 9a is made from two
200 mm X 30 mm boards. If <Tallow = 12 MPa and 'Tallow = 0.8 MPa,
determine if the beam can safely support the loading shown. Also, specify
the maximum spacing of nails needed to hold the two boards together if
each nail can safely resist 1.50 kN in shear.
1.5 kN
0.5 kN/m
c
- -2m - --1 - - -2 m - - - 1
30mm
(a)
SOLUTION
1.5 kN
The reactions on the beam are shown,
and the shear and moment diagrams are drawn in Fig. 15- 9b. Here
Vmax = 1.5 kN, Mmax = 2 kN · m.
Shear and Moment Diagrams.
2m - -
2m - -
1 kN
1.5 kN
The neutral axis (centroid) will be located from the
bottom of the beam. Working in units of meters, we have
Bending Stress.
I
V(kN)
1.5
2- A
y- = -y1------+----~ x
.___ __.. _ 1
M (kN·m)
I
2
IA
(m)
=
(0.1 m)(0.03 m)(0.2 m) + 0.215 m(0.03 m)(0.2 m)
0.03 m(0.2 m) + 0.03 m(0.2 m)
0.1575 m
Thus,
I = [
~
(0.03 m)(0.2 m) 3 + (0.03 m)(0.2 m)(0.1575 m 1
+[
(b)
Fig. 15-9
=
=
Since c
0.1 m)2 ]
1
(0.2 m)(0.03 m)3 + (0.03 m)(0.2 m)(0.215 m - 0.1575 m) 2 ]
12
60.125(10-6) m4
=
0.1575 m (not 0.230 m - 0.1575 m
12(106)
2(103) N · m(0.1575 m)
Pa ;::::
60.125(10-6) m4
=
=
0.0725 m), we require
5.24(106) Pa
OK
15.2
PRISMATIC BEAM DESIGN
7 09
Shear Stress. Maximum shear stress in the beam depends upon the
magnitude of Q and t. It occurs at the neutral axis, since Q is a maximum
there and at the neutral axis the thickness t = 0.03 m is the smallest for
the cross section. For simplicity, we will use the rectangular area below
the neutral axis to calculate Q, rather than a two-part composite area
above this axis, Fig. 15- 9c. We have
Q
=
y' A'
=
( 0 · 15~ 5 m )((0.1575 m)(0.03 m)] =
0.372(10- 3) m3
so that
g
To.0725m
N ---'o--f---+---- A
800 (
)
103
1.5(103) N(0.372(10- 3)] m3
Pa ;:-;::
60.125(10- 6) m4 (0.03 m)
309
=
(
)
103
LJ _ J.1575m
Pa
OK
-l
~
0.03m
Nail Spacing. From the shear diagram it is seen that the shear varies over
the entire span. Since the nail spacing depends on the magnitude of shear in
the beam, for simplicity (and to be conservative), we will design the spacing
on the basis of V = 1.5 kN for region BC, and V = 1 kN for region CD.
Since the nails join the flange to the web, Fig. 15- 9d, we have
Q
=
y' A'
=
(0.0725 m - 0.015 m)((0.2 m)(0.03 m)]
=
(c)
0.345(10- 3) m3
I a.2m I
The shear flow for each region is therefore
V8 cQ
qsc
q
CD
=
=
I
Vc0 Q
[
=
=
1.5(103) N(0.345(10- 3) m3]
60.125(10-6) m4
1(103) N(0.345(10- 3) m3]
60.125(10-6) m4
J_
=
N
8.61 kN/m
=
574kN/m
.
sco
=
1.50 kN
_ kN/m
8 61
=
1.50 kN
5.74 kN/m
=
0.174 m
=
0.261 m
I
A
LJ
(d)
One nail can resist 1.50 kN in shear, so the maximum spacing becomes
ssc
ITl-
O~m '--TfT-' [ 0.0725 m
For ease of measuring, use
s8 c
=
150mm
Ans.
sco
=
250mm
Ans.
Fig.15-9 (cont.)
710
15
CHAPTER
DESIGN OF BEAMS AND S HA FTS
FUNDAMENTAL PROBLEMS
Fl5-l. Determine the minimum dimension a to the nearest
mm of the beam's cross section to safely support the load.
The wood has an allowable normal stress of uauow = 10 MPa
and an allowable shear stress of Tallow = 1 MPa.
6kN
!
CTII2a
la-4
FlS-4. Determine the minimum dimension h to the nearest
k in. of the beam's cross section to safely support the load.
The wood has an allowable normal stress of uauow = 2 ksi
and an allowable shear stress of TaJJow = 200 psi.
6 kN
!
I·
lm~
6 ft
CI]}
lm
1-4 in.1
Prob.Fl5-l
Prob.FlS-4
FlS-2. Determine the minimum diameter d to the nearest
l in. of the rod to safely support the load. The rod is made of
a material having an allowable normal stress of uauow = 20 ksi
and an allowable shear stress of Tallow = 10 ksi.
Fl5-5. Determine the minimum dimension b to the nearest
mm of the beam's cross section to safely support the load.
The wood has an allowable normal stress of uauow = 12 MPa
and an allowable shear stress of TaJJow = 1.5 MPa.
SOkN
3 kip·ft
'Ill;
I •II
SkN·m
10 II \
11
1.5 ft
•
5 kN·m
0
~-1 m~l-1 m-f-~
1.5 ft
3 kip
m- l -1
Prob.Fl5-2
m-1 )
-I
3b
_l
lb l
Fl5-3. Determine the minimum dimension a to the nearest
mm of the beam's cross section to safely support the load.
The wood has an allowable normal stress of uauow = 12 MPa
and an allowable shear stress of Tallow = 1. 5 MP a.
I_
AJ'
j
"f
1-o.s m
Prob.Fl5-5
F15-6. Select the lightest W410·shaped section that can
I
In
1
m---11
safely support the load. The beam is made of steel having an
allowable normal stress of uauow = 150 MPa and an
allowable shear stress of Tauow = 75 MPa. Assume the beam
is pinned at A and roller supported at B.
~~~!
lSOkN
i -- - 2
Prob.Fl5-3
m
Prob.Fl5-6
1
m--1
15.2
711
PRISMATIC B EAM D ESIGN
PROBLEMS
15-1. The beam is made of timber that has an allowable
bending stress of u allow = 6.5 MPa and an allowable shear
stress of r allow = 500 kPa. Determine its dimensions if it is to
be rectangular and have a height-to-width ratio of 1.25.
Assume the beam rests on smooth supports.
15-5. Select the lightest-weight wide-flange beam from
Appendix B that will safely support the machine loading
shown. The allowable bending stress is uallow = 24 ksi and
the allowable shear stress is 'Tallow = 14 ksi.
5 kip
5 kip
5 kip
5 kip
8kN/m
I
I
I
I
I
I
I
M
l-2f1 ~-2ft -l-2f1 -l-2f1 -~2ft -I
-="=-
Prob.15-5
Prob.15-1
15-2. Determine the minimum width of the beam to the
nearest ~in. that will safely support the loading of P = 8 kip.
The allowable bending stress is u allow = 24 ksi and the
allowable shear stress is r allow = 15 ksi.
15-6. The spreader beam AB is used to slowly lift the
3000-lb pipe that is centrally located on the straps at C
and D. If the beam is a W12 x 45, determine if it can safely
support the load. The allowable bending stress is
u allow = 22 ksi and the allowable shear stress is Tallow = 12 ksi.
30001b
15-3. Solve Prob. 15-2 if P = 10 kip.
p
_ ! - - -6 ft - - - - - - 6 ft - - -1
~q~A
6inl_I
as
Probs. 15- 2/3
c
*15-4. The brick wall exerts a uniform distributed load of
1.20 kip/ft on the beam. If the allowable bending stress is
u allow = 22 ksi and the allowable shear stress is Tallow = 12 ksi,
select the lightest wide-flange section with the shortest depth
from Appendix B that will safely support the load. If there
are several choices of equal weight, choose the one with the
shortest height.
l ~ l.20 kip/ft
I +#~ l
D
Prob.15-6
15-7. Select the lightest-weight wide-flange beam with the
shortest depth from Appendix B that will safely support the
loading shown. The allowable bending stress is u allow = 24 ksi
and the allowableshearstressofrauow = 14ksi.
8 kip/ft
!!
~ ~ +t
l(.\
I
1-4 ft - 1 - -10 ft---~6 ft ~I '
- - - - - - - 6 ft - - - - - - --
Prob.15-4
Prob.15-7
•
1
1
712
CHAPTER
15
DESIGN OF BEAMS AND S HA FTS
*15-8. Select the lightest-weight wide-flange beam from
Appendix B that will safely support the loading shown. The
allowable bending stress u allow = 24 ksi and the allowable
shear stress of Tallow = 14 ksi.
15-11. The beam is constructed from two boards. If each
nail can support a shear force of 200 lb, determine the
maximum spacing of the nails, s, s' , and s", to the nearest
Ainch for regions AB, BC, and CD, respectively.
:r
SOOJb
15
r p
5 kip/ft
~
F
~ 6ft -~ 9ft
i
l_l- 8 in.-1
lin.-1
6 in.
L_
1500 lb
-tt1 in.
l~I
s"
1--'--l
C
D
- - -s ft - 1 -s ft - I
Prob.15-11
I
*15-12. The joists of a floor in a warehouse are to be
selected using square timber beams made of oak. If each
beam is to be designed to carry 90 lb/ft over a simply
supported span of 25 ft, determine the dimension a of its
square cross section to the nearest l in. The allowable
bending stress is u allow = 4.5 ksi and the allowable shear
stress is Tallow = 125 psi.
Prob.15-8
15-9. Select the lightest W360 wide-flange beam from
Appendix B that can safely support the loading. The beam
has an allowable normal stress of u allow = 150 MPa and an
allowable shear stress of Tallow = 80 MPa. Assume there is a
pin at A and a roller support at B.
15-10. Investigate if the W250 x 58 beam can safely
support the loading. The beam has an allowable normal
stress of u allow = 150 MPa and an allowable shear stress of
Tauow = 80 MPa. Assume there is a pin at A and a roller
support at B.
Prob.15-12
15-13. The timber beam has a width of 6 in. Determine its
height h so that it simultaneously reaches its allowable
bending stress u allow = 1.50 ksi and an allowable shear
stress of Tallow = 50 psi. Also, what is the maximum load P that
the beam can then support?
40 kN/m
0
,_A_ _ _ 4 m
8
- - - - + 11+--
Probs. 15-9/10
2m
50kN
p
!
~
-I
Prob.15-13
15.2
15-14. The beam is constructed from four boards. If each
nail can support a shear force of 300 lb, determine the
maximum spacing of the nails. s, s' ands•, for regions AB,
BC, and CD. respectively.
s
8
A
I,_ _
6 ft
s"
s'
1-1
f--1
--'~-~--- C.....:::~1::=6 ft
6 ft
BEAM
DESIGN
713
*15-16. If the cable is subjected to a maximum force of
P = 50 kN, select the lightest W310 wide-flange beam that
can safely support the load. The beam has an allowable
normal stress of u allow = 150 MPa and an allowable shear
stress of Tallow = 85 M Pa.
15-17. If the W360 x 45 wide-flange beam has an allowable
normal stress of Uano.. = 150 MPa and an allowable shear
stress of T anow = 85 MPa. determine the maximum cable force
P that can safely be supported by the beam.
!
1 kip
3 kip
PRISMATIC
I
-I
D
lin.-rD
,L1- 9 in. -1
1- + - - - 2 m
7 in.
•
1 Ill
p
_L
T-111 in.
2 m - -+-1
- 11l in.
Probs. 15-16/17
Prob. 15-14
15-15. The beam is constructed from two boards. If each
nail can support a shear force of 200 lb, determine the
maximum spacing of the nails, s. s'. and s•. to the nearest
in. for regions AB, BC. and CD. respectively.
l
15-18. If P = 800 lb, determine the minimum dimension
a of the beam's cross section to the nearest ~in. to safely
support the load. The wood has an allowable normal stress
of u allow = 1.5 ksi and an allowable shear stress of
Tallow = 150 psi.
15-19. If a = 3 in. and the wood has an allowable normal
stress of u a11o.. = 1.5 ksi. and an allowable shear stress of
T a11ow = 150 psi, determine the maximum allowable value
of P that c